Deck 13: Boundary-Value Problems in Other Coordinate Systems

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سؤال
In the previous problem, the solution of the eigenvalue problem is

A) λ=nπ,T=ancos(nπi)+bnsin(nπt)\lambda = n \pi , T = a _ { n } \cos ( n \pi i ) + b _ { n } \sin ( n \pi t )
B) λ=n2π2,T=ancos(nπt)+bnsin(nπt)\lambda = n ^ { 2 } \pi ^ { 2 } , T = a _ { n } \cos ( n \pi t ) + b _ { n } \sin ( n \pi t )
C) λ=zn2/4\lambda = z _ { n } ^ { 2 } / 4 , where J0(zn)=0,R=J0(znr/2)J _ { 0 } \left( z _ { n } \right) = 0 , R = J _ { 0 } \left( z _ { n } r / 2 \right)
D) λ=zn/2\lambda = z _ { n } / 2 , where J0(zn)=0,R=J0(znr/2)J _ { 0 } \left( z _ { n } \right) = 0 , R = J _ { 0 } \left( z _ { n } r / 2 \right)
E) λ=zn2\lambda = z _ { n } ^ { 2 } , where J0(zn)=0,R=J0(znr)J _ { 0 } \left( z _ { n } \right) = 0 , R = J _ { 0 } \left( z _ { n } r \right)
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سؤال
Using the separation of the previous problem, the equation for Θ\Theta becomes

A) ΘλΘ=0\Theta ^ { \prime \prime } - \lambda \Theta = 0
B) Θ+λΘ=0\Theta ^ { \prime \prime } + \lambda \Theta = 0
C) sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0
D) cos(θ)Θsin(θ)Θ+λsin(θ)Θ=0\cos ( \theta ) \Theta ^ { \prime \prime } - \sin ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0
E) sin(θ)Θcos(θ)Θ+λsin(θ)Θ=0\sin ( \theta ) \Theta ^ { \prime \prime } - \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0
سؤال
The two dimensional wave equation in polar coordinates is

A) a2(2yr21rur+1r22uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
B) a2(2yr2+1rur1r22uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
C) a2(2yr2+1rur+1r22uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
D) a2(2yr2+1r2ur+1r2uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
E) a2(2yr2+1r2ur1r2uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } - \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
سؤال
Consider the vibrations of a circular membrane of radius 2 clamped along the circumference, with an initial displacement of 1sin(πr/2)1 - \sin ( \pi r / 2 ) and an initial velocity of zero. The mathematical model for this situation is

A) 2ur2+1rur=2ut2=0,u(2,t)=0,u(r,0)=1sin(πr/2),u,(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi r / 2 ) , u , ( r , 0 ) = 0
B) 2ur21rur=2ut2=0,u(2,t)=0,u(r,0)=1sin(πr/2),u,(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi r / 2 ) , u , ( r , 0 ) = 0
C) 2ur2+1rur=2ut2=0,u(2,t)=0,u(r,0)=1sin(π/2),ut(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } = - \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi / 2 ) , u _ { t } ( r , 0 ) = 0
D) 2ur21r2ur=2ut2=0,u(2,t)=0,u(r,0)=1sin(πr/2),ut(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } = - \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi r / 2 ) , u _ { t } ( r , 0 ) = 0
E) 2ur2+1r2ur=2ut2=0,u(2,t)=0,u(r,0)=1sin(πr/2),u,(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi r / 2 ) , u , ( r , 0 ) = 0
سؤال
In the previous four problems, the infinite series solution of the original problem is (for certain values of the constants ana _ { n } and bnb _ { n } )

A) u=n=1anJ0(znr/2)cos(znt/2)u = \sum _ { n = 1 } ^ { \infty } a _ { n } J _ { 0 } \left( z _ { n } r / 2 \right) \cos \left( z _ { n } t / 2 \right)
B) u=n=1bnJ0(znr/2)sin(znt/2)u = \sum _ { n = 1 } ^ { \infty } b _ { n } J _ { 0 } \left( z _ { n } r / 2 \right) \sin \left( z _ { n } t / 2 \right)
C) u=n=1bnJ0(nπr)sin(nπt)u = \sum _ { n = 1 } ^ { \infty } b _ { n } J _ { 0 } ( n \pi r ) \sin ( n \pi t )
D) u=n=1anJ0(nπr)cos(nπt)u = \sum _ { n = 1 } ^ { \infty } a _ { n } J _ { 0 } ( n \pi r ) \cos ( n \pi t )
E) u=n1J0(znr/2)(ancos(nπt)+bnsin(nπt))u = \sum _ { n - 1 } ^ { \infty } J _ { 0 } \left( z _ { n } r / 2 \right) \left( a _ { n } \cos ( n \pi t ) + b _ { n } \sin ( n \pi t ) \right)
سؤال
The equations relating Cartesian and spherical coordinates include Select all that apply.

A) y=rcosϕcosθy = r \cos \phi \cos \theta
B) x=rcosϕsinθx = r \cos \phi \sin \theta
C) r2=x2+y2+z2r ^ { 2 } = x ^ { 2 } + y ^ { 2 } + z ^ { 2 }
D) z=rcosθz = r \cos \theta
E) tanϕ=y/x\tan \phi = y / x
سؤال
In the previous problem, separate variables using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems are

A) R+rRr2λR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0R ^ { \prime \prime } + r R ^ { \prime } - r ^ { 2 } \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
B) R+rR+r2λR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0R ^ { \prime \prime } + r R ^ { \prime } + r ^ { 2 } \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
C) r2R+rRλR=0,ΘλΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } - \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
D) r2R+rR+λR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
E) r2R+rRλR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
سؤال
The solution of r2R+rR+λR=0,R(1)=0,R(2)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , R ( 1 ) = 0 , R ( 2 ) = 0 is

A) λ=nπ/ln2,R=sin(nπlnr/ln2)\lambda = n \pi / \ln 2 , R = \sin ( n \pi \ln r / \ln 2 )
B) λ=(nπ/ln2)2,R=sin(nπlnr/ln2)\lambda = ( n \pi / \ln 2 ) ^ { 2 } , R = \sin ( n \pi \ln r / \ln 2 )
C) λ=(nπ/ln2)2,R=cos(nπlnr/ln2)\lambda = ( n \pi / \ln 2 ) ^ { 2 } , R = \cos ( n \pi \ln r / \ln 2 )
D) λ=nπ/ln2,R=cos(nπlnr/ln2)\lambda = n \pi / \ln 2 , R = \cos ( n \pi \ln r / \ln 2 )
E) none of the above
سؤال
In the previous four problems, the infinite series solution of the original problem is u=n=1rn(Ancos(nθ)+Bnsin(nθ))u = \sum _ { n = 1 } ^ { \infty } r ^ { n } \left( A _ { n } \cos ( n \theta ) + B _ { n } \sin ( n \theta ) \right) where Select all that apply.

A) An=02πf(θ)cos(nθ)dθ/(cnπ)A _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \cos ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
B) Bn=02πf(θ)sin(nθ)dθ/(cnπ)B _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \sin ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
C) Bn=02πf(θ)cos(nθ)dθ/(cnπ)B _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \cos ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
D) An=0A _ { n } = 0
E) Bn=0B _ { n } = 0
سؤال
In changing from Cartesian to spherical coordinates, rx\frac { \partial r } { \partial x } becomes

A) sinϕsinθ\sin \phi \sin \theta
B) sinϕcosθ\sin \phi \cos \theta
C) cosϕsinθ\cos \phi \sin \theta
D) cosϕcosθ\cos \phi \cos \theta
E) sinϕ\sin \phi
سؤال
In the three previous problems, the product solutions are

A) un=rn(cnenθ+dnenθ),n=0,1,2,u _ { n } = r ^ { n } \left( c _ { n } e ^ { n \theta } + d _ { n } e ^ { - n \theta } \right) , n = 0,1,2 , \ldots
B) un=rn(cnenθ+dnenθ),n=1,2,3,u _ { n } = r ^ { n } \left( c _ { n } e ^ { n \theta } + d _ { n } e ^ { - n \theta } \right) , n = 1,2,3 , \ldots
C) un=rn(cncos(nθ)+dnsin(nθ)),n=0,1,2,u _ { n } = r ^ { n } \left( c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) \right) , n = 0,1,2 , \ldots
D) un=rncos(nθ),n=0,1,2,u _ { n } = r ^ { n } \cos ( n \theta ) , n = 0,1,2 , \ldots
E) un=rnsin(nθ),n=1,2,3,u _ { n } = r ^ { n } \sin ( n \theta ) , n = 1,2,3 , \ldots
سؤال
In the previous three problems, the solution for RR is

A) R=rnR = r ^ { n }
B) R=rn1R = r ^ { - n - 1 }
C) R=c1rn+c2rn1R = c _ { 1 } r ^ { n } + c _ { 2 } r ^ { - n - 1 }
D) R=c1rn+c2rn+1R = c _ { 1 } r ^ { n } + c _ { 2 } r ^ { n + 1 }
E) R=c1rn+c2rn1R = c _ { 1 } r ^ { n } + c _ { 2 } r ^ { n - 1 }
سؤال
In the previous problem, the solution of the eigenvalue problem is

A) λ=n2,R=rn,n=0,1,2,\lambda = n ^ { 2 } , R = r ^ { n } , n = 0,1,2 , \ldots
B) λ=n,R=rn,n=0,1,2,\lambda = n , R = r ^ { n } , n = 0,1,2 , \ldots
C) λ=n,Θ=cncos(nθ)+dnsin(nθ),n=0,1,2,.\lambda = n , \Theta = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , .
D) λ=n2,Θ=dnsin(nθ),n=1,2,3,.\lambda = n ^ { 2 } , \Theta = d _ { n } \sin ( n \theta ) , n = 1,2,3 , .
E) λ=n2,Θ=cncos(nθ),n=0,1,2,\lambda = n ^ { 2 } , \Theta = c _ { n } \cos ( n \theta ) , n = 0,1,2 , \ldots
سؤال
In separating variables, using, u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) , in the equation 2ur2+2rur+1r22uθ2+cotθr2uθ=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 2 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } + \frac { \cot \theta } { r ^ { 2 } } \frac { \partial u } { \partial \theta } = 0 , the resulting equation for RR and is

A) r2R+2rRλR=0r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda R = 0
B) r2R2rRλR=0r ^ { 2 } R ^ { \prime \prime } - 2 r R ^ { \prime } - \lambda R = 0
C) rR+2rRλrR=0r R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda r R = 0
D) rR+2RλrR=0r R ^ { \prime \prime } + 2 R ^ { \prime } - \lambda r R = 0
E) r2R2rRλrR=0r ^ { 2 } R ^ { \prime \prime } - 2 r R ^ { \prime } - \lambda r R = 0
سؤال
In the previous problem, if we also require that Θ\Theta be bounded everywhere, the solution of the eigenvalue problem is

A) λ=n(n+1),Θ=Pn(sinθ),n=1,2,3,\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \sin \theta ) , n = 1,2,3 , \ldots
B) λ=n(n1),Θ=sin(nθ),n=1,2,3,\lambda = n ( n - 1 ) , \Theta = \sin ( n \theta ) , n = 1,2,3 , \ldots
C) λ=n(n1),Θ=cos(nθ),n=1,2,3,\lambda = n ( n - 1 ) , \Theta = \cos ( n \theta ) , n = 1,2,3 , \ldots
D) λ=n(n+1),Θ=Pn(cosθ),n=1,2,3,\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \cos \theta ) , n = 1,2,3 , \ldots
E) λ=n(n1),Θ=Pn(sinθ),n=1,2,3,\lambda = n ( n - 1 ) , \Theta = P _ { n } ( \sin \theta ) , n = 1,2,3 , \ldots
سؤال
In changing from Cartesian to polar coordinates, ry\frac { \partial r } { \partial y } is

A) sinθ\sin \theta
B) cosθ\cos \theta
C) sinθ/r\sin \theta / r
D) cosθ/r\cos \theta / r
E) sinθ/r- \sin \theta / r
سؤال
In changing from Cartesian to polar coordinates, uy\frac { \partial u } { \partial y } is

A) cosθursinθuθ/r\cos \theta \frac { \partial u } { \partial r } - \sin \theta \frac { \partial u } { \partial \theta } / r
B) cosθur+cosθuθ/r\cos \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta } / r
C) sinθur+cosθuθ\sin \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta }
D) sinθur+cosθuθ/r\sin \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta } / r
E) sinθurcosθuθ/r\sin \theta \frac { \partial u } { \partial r } - \cos \theta \frac { \partial u } { \partial \theta } / r
سؤال
In the previous problem, after separating variables using u(r,t)=R(r)T(t)u ( r , t ) = R ( r ) T ( t ) . the resulting problems are

A) rR+R+rλR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + r \lambda R = 0 , R ( 0 ) is bounded R(2)=0,TλT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } - \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
B) rR+R+rλR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + r \lambda R = 0 , R ( 0 ) is bounded R(2)=0,T+λT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
C) rR+RrλR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } - r \lambda R = 0 , R ( 0 ) is bounded R(2)=0,T+λT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
D) rR+R+λR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + \lambda R = 0 , R ( 0 ) is bounded R(2)=0,T+λT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
E) rR+R+λR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + \lambda R = 0 , R ( 0 ) is bounded R(2)=0,TλT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } - \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
سؤال
Consider the steady-state temperature distribution in a circular disc of radius CC centere at the origin, with temperature given as a function, f(θ)f ( \theta ) on the boundary r=cr = c and zero on the boundaries θ=0\theta = 0 and θ=π\theta = \pi The mathematical model of this situation is

A) 2ur21rur+1r22uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
B) 2ur2+1rur1r22uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
C) 2ur21rur1r22uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
D) 2ur2+1r2ur+1r2uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
E) 2ur2+1rur+1r22uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
سؤال
In the previous three problems, the product solutions are

A) u=J0(znr/2)(ancos(znt/2)+bnsin(znt/2))u = J _ { 0 } \left( z _ { n } r / 2 \right) \left( a _ { n } \cos \left( z _ { n } t / 2 \right) + b _ { n } \sin \left( z _ { n } t / 2 \right) \right)
B) u=J0(znr/2)ancos(znt/2)u = J _ { 0 } \left( z _ { n } r / 2 \right) a _ { n } \cos \left( z _ { n } t / 2 \right)
C) u=J0(znr/2)bnsin(znt/2)u = J _ { 0 } \left( z _ { n } r / 2 \right) b _ { n } \sin \left( z _ { n } t / 2 \right)
D) u=J0(nπr)bnsin(nπi)u = J _ { 0 } ( n \pi r ) b _ { n } \sin ( n \pi i )
E) u=J0(nπr)ancos(nπi)u = J _ { 0 } ( n \pi r ) a _ { n } \cos ( n \pi i )
سؤال
In the previous problem, the solution of the eigenvalue problem is

A) λ=n,Θ=cncos(nθ)+dnsin(nθ),n=0,1,2,\lambda = n , \Theta = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots
B) λ=n2,=cncos(nθ)+dnsin(nθ),n=0,1,2,\lambda = n ^ { 2 } , \ominus = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots
C) λ=n2,=cncos(nθ)+dnsin(nθ),n=1,2,3,\lambda = n ^ { 2 } , \ominus = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 1,2,3 , \ldots
D) λ=n2,R=rn,n=0,1,2,\lambda = n ^ { 2 } , R = r ^ { n } , n = 0,1,2 , \ldots
E) λ=n,R=rn,n=0,1,2,\lambda = n , R = r ^ { n } , n = 0,1,2 , \ldots
سؤال
In changing from Cartesian to polar coordinates, rx\frac { \partial r } { \partial x } is

A) sinθ\sin \theta
B) cosθ\cos \theta
C) sinθ/r\sin \theta / r
D) cosθ/r\cos \theta / r
E) sinθ/r- \sin \theta / r
سؤال
In the previous problem, separate variables using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems are

A) r2R+rRλR=0,ΘλΘ=0,Θ(θ)=Θ(θ+2π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } - \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
B) r2R+rR+λR=0,Θ+λΘ=0,Θ(θ)=Θ(θ+2π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
C) r2R+rRλR=0,Θ+λΘ=0,Θ(θ)=Θ(θ+2π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
D) r2R+rRλR=0,Θ+λΘ=0,Θ(θ)=Θ(θ+π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + \pi )
E) r2R+rR+λR=0,Θ+λΘ=0,Θ(θ)=Θ(θ+π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + \pi )
سؤال
In changing from Cartesian to polar coordinates, ux\frac { \partial u } { \partial x } is

A) sinθur+cosθuθ/r\sin \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta } / r
B) sinθurcosθuθ/r\sin \theta \frac { \partial u } { \partial r } - \cos \theta \frac { \partial u } { \partial \theta } / r
C) cosθursinθuθ/r\cos \theta \frac { \partial u } { \partial r } - \sin \theta \frac { \partial u } { \partial \theta } / r
D) cosθur+cosθuθ/r\cos \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta } / r
E) sinθur+cosθuθ\sin \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta }
سؤال
The relationships between Cartesian and spherical coordinates are Select all that apply.

A) r2=x2+y2r ^ { 2 } = x ^ { 2 } + y ^ { 2 }
B) tanϕ=y/x\tan \phi = y / x
C) x=rcosϕsinθx = r \cos \phi \sin \theta
D) y=rcosϕcosθy = r \cos \phi \cos \theta
E) z=rcosθz = r \cos \theta
سؤال
In the previous three problems, the product solutions are

A) u=J0(nπr/3)cos(nπz/3)u = J _ { 0 } ( n \pi r / 3 ) \cos ( n \pi z / 3 )
B) u=J0(n2π2r/9)cos(nπz/3)u = J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \cos ( n \pi z / 3 )
C) u=J0(n2π2r/9)sin(nπz/3)u = J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \sin ( n \pi z / 3 )
D) u=J0(znr)cosh(zn(z3))u = J _ { 0 } \left( z _ { n } r \right) \cosh \left( z _ { n } ( z - 3 ) \right)
E) u=J0(znr/2)sinh(zn(z3)/2)u = J _ { 0 } \left( z _ { n } r / 2 \right) \sinh \left( z _ { n } ( z - 3 ) / 2 \right)
سؤال
Consider the steady-state temperature distribution in a circular cylinder of radius 2 and height 3, with zero temperature at r=2r = 2 and at z=3z = 3 and a temperature of 10 at z=0z = 0 . The mathematical model for this problem is

A) 2ur2+1rur2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
B) 2ur21rur2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
C) 2ur21r2ur+2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
D) 2ur2+1rur+2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
E) 2ur2+1r2ur+2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
سؤال
In the problem 2ur2+2rur+1r22uθ2+cotθr2uθ=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 2 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } + \frac { \cot \theta } { r ^ { 2 } } \frac { \partial u } { \partial \theta } = 0 , separate variables, using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems for RR and Θ\Theta are

A) r2R+2rR+λR=0,R(0)r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } + \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
B) r2R+2rRλR=0,R(0)r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
C) r2R+2rRλR=0,R(0)r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θ+cos(θ)Θλsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } - \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
D) r2R2rRλR=0,R(0)r ^ { 2 } R ^ { \prime \prime } - 2 r R ^ { \prime } - \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
E) r2R+2rRλR=0,R(0)r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θcos(θ)Θ+λsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } - \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
سؤال
The Laplacian in polar coordinates is

A) 2ur2+1r2ur+1r2uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
B) 2ur2+1rur+1r22uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
C) 2ur21rur+1r22uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
D) 2ur2+1rur1r22uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
E) 2ur21rur1r22uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
سؤال
Consider the steady-state temperature distribution in a circular disc of radius CC centere at the origin, with temperature given as a function, f(θ)f ( \theta ) on the boundary. The mathematical model of this situation is

A) 2ur21rur+1r22uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
B) 2ur2+1rur1r22uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
C) 2ur21rur1r22uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
D) 2ur2+1r2ur+1r2uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
E) 2ur2+1rur+1r22uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
سؤال
In the previous four problems, the infinite series solution is (for certain values of the constants)

A) u=n1cnJ0(nπr/3)cos(nπz/3)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } ( n \pi r / 3 ) \cos ( n \pi z / 3 )
B) u=n1cnJ0(n2π2r/9)cos(nπz/3)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \cos ( n \pi z / 3 )
C) u=n1cnJ0(n2π2r/9)sin(nπz/3)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \sin ( n \pi z / 3 )
D) u=n1cnJ0(znr/2)sinh(zn(z3)/2)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( z _ { n } r / 2 \right) \sinh \left( z _ { n } ( z - 3 ) / 2 \right)
E) u=n=1cnJ0(znr)cosh(zn(z3))u = \sum _ { n = 1 } ^ { \infty } c _ { n } J _ { 0 } \left( z _ { n } r \right) \cosh \left( z _ { n } ( z - 3 ) \right)
سؤال
In the three previous problems, the product solutions are

A) un=(anrn+bnrn)(cncos(nθ)+dnsin(nθ)),n=0,1,2,u _ { n } = \left( a _ { n } r ^ { n } + b _ { n } r ^ { - n } \right) \left( c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) \right) , n = 0,1,2 , \ldots
B) un=rn(cncos(nθ)+dnsin(nθ)),n=1,2,3,u _ { n } = r ^ { n } \left( c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) \right) , n = 1,2,3 , \ldots
C) un=rn(cncos(nθ)+dnsin(nθ)),n=0,1,2,u _ { n } = r ^ { n } \left( c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) \right) , n = 0,1,2 , \ldots
D) un=rn(cnenθ+dnenθ),n=0,1,2,u _ { n } = r ^ { n } \left( c _ { n } e ^ { n \theta } + d _ { n } e ^ { - n \theta } \right) , n = 0,1,2 , \ldots
E) un=rn(cnenθ+dnenθ),n=1,2,3,u _ { n } = r ^ { n } \left( c _ { n } e ^ { n \theta } + d _ { n } e ^ { - n \theta } \right) , n = 1,2,3 , \ldots
سؤال
The solution of r2R+rR+λR=0,R(0)=0,R(1)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , R ( 0 ) = 0 , R ( 1 ) = 0 is

A) λ=nπ,R=sin(nπlnr)\lambda = n \pi , R = \sin ( n \pi \ln r )
B) λ=(nπ)2,R=sin(nπlnr)\lambda = ( n \pi ) ^ { 2 } , R = \sin ( n \pi \ln r )
C) λ=(nπ)2,R=sin(nπlnr)+cos(nπlnr)\lambda = ( n \pi ) ^ { 2 } , R = \sin ( n \pi \ln r ) + \cos ( n \pi \ln r )
D) λ=nπ,R=sin(nπlnr)cos(nπlnr)\lambda = n \pi , R = \sin ( n \pi \ln r ) - \cos ( n \pi \ln r )
E) none of the above
سؤال
In the previous problem, the solution of the eigenvalue problem is

A) λ=n(n+1),R=rn,n=1,2,3,\lambda = n ( n + 1 ) , R = r ^ { n } , n = 1,2,3 , \ldots
B) λ=n(n1),R=rn,n=1,2,3,\lambda = n ( n - 1 ) , R = r ^ { n } , n = 1,2,3 , \ldots
C) λ=n(n1),Θ=Pn(cosθ),n=1,2,3,\lambda = n ( n - 1 ) , \Theta = P _ { n } ( \cos \theta ) , n = 1,2,3 , \ldots
D) λ=n(n+1),Θ=Pn(cosθ),n=1,2,3,\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \cos \theta ) , n = 1,2,3 , \ldots
E) λ=n(n+1),Θ=Pn(sinθ),n=1,2,3,\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \sin \theta ) , n = 1,2,3 , \ldots
سؤال
In the previous two problems, the product solutions are

A) un=rnPn(cosθ)u _ { n } = r ^ { n } P _ { n } ( \cos \theta )
B) un=rnPn(cosθ)u _ { n } = r ^ { - n } P _ { n } ( \cos \theta )
C) un=rnPn(sinθ)u _ { n } = r ^ { n } P _ { n } ( \sin \theta )
D) un=rnPn(sinθ)u _ { n } = r ^ { - n } P _ { n } ( \sin \theta )
E) none of the above
سؤال
In the previous problem, after separating variables, the resulting problems are

A) rRR+λrR=0,R(0)r R ^ { \prime \prime } - R ^ { \prime } + \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,Z+λZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } + \lambda Z = 0 , Z ( 3 ) = 0
B) rR+RλrR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } - \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,ZλZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } - \lambda Z = 0 , Z ( 3 ) = 0
C) rRRλrR=0,R(0)r R ^ { \prime \prime } - R ^ { \prime } - \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,ZλZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } - \lambda Z = 0 , Z ( 3 ) = 0
D) rR+R+λrR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,ZλZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } - \lambda Z = 0 , Z ( 3 ) = 0
E) rR+R+λrR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,Z+λZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } + \lambda Z = 0 , Z ( 3 ) = 0
سؤال
When changing from Cartesian to spherical coordinates, ϕx=\frac { \partial \phi } { \partial x } =

A) sinϕ/(rsinθ)- \sin \phi / ( r \sin \theta )
B) sinϕ/(rsinθ)\sin \phi / ( r \sin \theta )
C) cosϕ/(rsinθ)- \cos \phi / ( r \sin \theta )
D) cosϕ/(rcosθ)\cos \phi / ( r \cos \theta )
E) sinϕ/(rcosθ)\sin \phi / ( r \cos \theta )
سؤال
In the previous four problems, the infinite series solution of the original problem is u=A0+n=1rn(Ancos(nθ)+Bnsin(nθ))u = A _ { 0 } + \sum _ { n = 1 } ^ { \infty } r ^ { n } \left( A _ { n } \cos ( n \theta ) + B _ { n } \sin ( n \theta ) \right) where Select all that apply.

A) A0=02πf(θ)dθl(2π)A _ { 0 } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) d \theta l ( 2 \pi )
B) An=02πf(θ)sin(nθ)dθ/(cnπ)A _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \sin ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
C) An=02πf(θ)cos(nθ)dθ/(cnπ)A _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \cos ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
D) Bn=02πf(θ)sin(nθ)dθ/(cnπ)B _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \sin ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
E) Bn=02πf(θ)cos(nθ)dθ/(cnπ)B _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \cos ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
سؤال
In the previous problem, the solution of the eigenvalue problem is

A) λ=zn2/4\lambda = z _ { n } ^ { 2 } / 4 , where J0(zn)=0;R=J0(znr/2)J _ { 0 } \left( z _ { n } \right) = 0 ; R = J _ { 0 } \left( z _ { n } r / 2 \right)
B) λ=zn2\lambda = z _ { n } ^ { 2 } , where J0(zn)=0;R=J0(znr)J _ { 0 } \left( z _ { n } \right) = 0 ; R = J _ { 0 } \left( z _ { n } r \right)
C) λ=nπ/9,Z=cos(nπz/3)\lambda = n \pi / 9 , Z = \cos ( n \pi z / 3 )
D) λ=n2π2/9,Z=cos(nπz/3)\lambda = n ^ { 2 } \pi ^ { 2 } / 9 , Z = \cos ( n \pi z / 3 )
E) λ=n2π2/9,Z=sin(nπz/3)\lambda = n ^ { 2 } \pi ^ { 2 } / 9 , Z = \sin ( n \pi z / 3 )
سؤال
When changing from Cartesian to spherical coordinates, rz=\frac { \partial r } { \partial z } =

A) cosϕ\cos \phi
B) cosθ\cos \theta
C) tanθ\tan \theta
D) sinϕ\sin \phi
E) sinθ\sin \theta
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Deck 13: Boundary-Value Problems in Other Coordinate Systems
1
In the previous problem, the solution of the eigenvalue problem is

A) λ=nπ,T=ancos(nπi)+bnsin(nπt)\lambda = n \pi , T = a _ { n } \cos ( n \pi i ) + b _ { n } \sin ( n \pi t )
B) λ=n2π2,T=ancos(nπt)+bnsin(nπt)\lambda = n ^ { 2 } \pi ^ { 2 } , T = a _ { n } \cos ( n \pi t ) + b _ { n } \sin ( n \pi t )
C) λ=zn2/4\lambda = z _ { n } ^ { 2 } / 4 , where J0(zn)=0,R=J0(znr/2)J _ { 0 } \left( z _ { n } \right) = 0 , R = J _ { 0 } \left( z _ { n } r / 2 \right)
D) λ=zn/2\lambda = z _ { n } / 2 , where J0(zn)=0,R=J0(znr/2)J _ { 0 } \left( z _ { n } \right) = 0 , R = J _ { 0 } \left( z _ { n } r / 2 \right)
E) λ=zn2\lambda = z _ { n } ^ { 2 } , where J0(zn)=0,R=J0(znr)J _ { 0 } \left( z _ { n } \right) = 0 , R = J _ { 0 } \left( z _ { n } r \right)
λ=zn2/4\lambda = z _ { n } ^ { 2 } / 4 , where J0(zn)=0,R=J0(znr/2)J _ { 0 } \left( z _ { n } \right) = 0 , R = J _ { 0 } \left( z _ { n } r / 2 \right)
2
Using the separation of the previous problem, the equation for Θ\Theta becomes

A) ΘλΘ=0\Theta ^ { \prime \prime } - \lambda \Theta = 0
B) Θ+λΘ=0\Theta ^ { \prime \prime } + \lambda \Theta = 0
C) sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0
D) cos(θ)Θsin(θ)Θ+λsin(θ)Θ=0\cos ( \theta ) \Theta ^ { \prime \prime } - \sin ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0
E) sin(θ)Θcos(θ)Θ+λsin(θ)Θ=0\sin ( \theta ) \Theta ^ { \prime \prime } - \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0
sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0
3
The two dimensional wave equation in polar coordinates is

A) a2(2yr21rur+1r22uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
B) a2(2yr2+1rur1r22uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
C) a2(2yr2+1rur+1r22uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
D) a2(2yr2+1r2ur+1r2uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
E) a2(2yr2+1r2ur1r2uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } - \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
a2(2yr2+1rur+1r22uθ2)=2ut2a ^ { 2 } \left( \frac { \partial ^ { 2 } y } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } \right) = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } }
4
Consider the vibrations of a circular membrane of radius 2 clamped along the circumference, with an initial displacement of 1sin(πr/2)1 - \sin ( \pi r / 2 ) and an initial velocity of zero. The mathematical model for this situation is

A) 2ur2+1rur=2ut2=0,u(2,t)=0,u(r,0)=1sin(πr/2),u,(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi r / 2 ) , u , ( r , 0 ) = 0
B) 2ur21rur=2ut2=0,u(2,t)=0,u(r,0)=1sin(πr/2),u,(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi r / 2 ) , u , ( r , 0 ) = 0
C) 2ur2+1rur=2ut2=0,u(2,t)=0,u(r,0)=1sin(π/2),ut(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } = - \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi / 2 ) , u _ { t } ( r , 0 ) = 0
D) 2ur21r2ur=2ut2=0,u(2,t)=0,u(r,0)=1sin(πr/2),ut(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } = - \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi r / 2 ) , u _ { t } ( r , 0 ) = 0
E) 2ur2+1r2ur=2ut2=0,u(2,t)=0,u(r,0)=1sin(πr/2),u,(r,0)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 2 , t ) = 0 , u ( r , 0 ) = 1 - \sin ( \pi r / 2 ) , u , ( r , 0 ) = 0
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In the previous four problems, the infinite series solution of the original problem is (for certain values of the constants ana _ { n } and bnb _ { n } )

A) u=n=1anJ0(znr/2)cos(znt/2)u = \sum _ { n = 1 } ^ { \infty } a _ { n } J _ { 0 } \left( z _ { n } r / 2 \right) \cos \left( z _ { n } t / 2 \right)
B) u=n=1bnJ0(znr/2)sin(znt/2)u = \sum _ { n = 1 } ^ { \infty } b _ { n } J _ { 0 } \left( z _ { n } r / 2 \right) \sin \left( z _ { n } t / 2 \right)
C) u=n=1bnJ0(nπr)sin(nπt)u = \sum _ { n = 1 } ^ { \infty } b _ { n } J _ { 0 } ( n \pi r ) \sin ( n \pi t )
D) u=n=1anJ0(nπr)cos(nπt)u = \sum _ { n = 1 } ^ { \infty } a _ { n } J _ { 0 } ( n \pi r ) \cos ( n \pi t )
E) u=n1J0(znr/2)(ancos(nπt)+bnsin(nπt))u = \sum _ { n - 1 } ^ { \infty } J _ { 0 } \left( z _ { n } r / 2 \right) \left( a _ { n } \cos ( n \pi t ) + b _ { n } \sin ( n \pi t ) \right)
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The equations relating Cartesian and spherical coordinates include Select all that apply.

A) y=rcosϕcosθy = r \cos \phi \cos \theta
B) x=rcosϕsinθx = r \cos \phi \sin \theta
C) r2=x2+y2+z2r ^ { 2 } = x ^ { 2 } + y ^ { 2 } + z ^ { 2 }
D) z=rcosθz = r \cos \theta
E) tanϕ=y/x\tan \phi = y / x
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In the previous problem, separate variables using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems are

A) R+rRr2λR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0R ^ { \prime \prime } + r R ^ { \prime } - r ^ { 2 } \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
B) R+rR+r2λR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0R ^ { \prime \prime } + r R ^ { \prime } + r ^ { 2 } \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
C) r2R+rRλR=0,ΘλΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } - \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
D) r2R+rR+λR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
E) r2R+rRλR=0,Θ+λΘ=0,Θ(0)=0,Θ(π)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( 0 ) = 0 , \Theta ( \pi ) = 0
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The solution of r2R+rR+λR=0,R(1)=0,R(2)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , R ( 1 ) = 0 , R ( 2 ) = 0 is

A) λ=nπ/ln2,R=sin(nπlnr/ln2)\lambda = n \pi / \ln 2 , R = \sin ( n \pi \ln r / \ln 2 )
B) λ=(nπ/ln2)2,R=sin(nπlnr/ln2)\lambda = ( n \pi / \ln 2 ) ^ { 2 } , R = \sin ( n \pi \ln r / \ln 2 )
C) λ=(nπ/ln2)2,R=cos(nπlnr/ln2)\lambda = ( n \pi / \ln 2 ) ^ { 2 } , R = \cos ( n \pi \ln r / \ln 2 )
D) λ=nπ/ln2,R=cos(nπlnr/ln2)\lambda = n \pi / \ln 2 , R = \cos ( n \pi \ln r / \ln 2 )
E) none of the above
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9
In the previous four problems, the infinite series solution of the original problem is u=n=1rn(Ancos(nθ)+Bnsin(nθ))u = \sum _ { n = 1 } ^ { \infty } r ^ { n } \left( A _ { n } \cos ( n \theta ) + B _ { n } \sin ( n \theta ) \right) where Select all that apply.

A) An=02πf(θ)cos(nθ)dθ/(cnπ)A _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \cos ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
B) Bn=02πf(θ)sin(nθ)dθ/(cnπ)B _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \sin ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
C) Bn=02πf(θ)cos(nθ)dθ/(cnπ)B _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \cos ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
D) An=0A _ { n } = 0
E) Bn=0B _ { n } = 0
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In changing from Cartesian to spherical coordinates, rx\frac { \partial r } { \partial x } becomes

A) sinϕsinθ\sin \phi \sin \theta
B) sinϕcosθ\sin \phi \cos \theta
C) cosϕsinθ\cos \phi \sin \theta
D) cosϕcosθ\cos \phi \cos \theta
E) sinϕ\sin \phi
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In the three previous problems, the product solutions are

A) un=rn(cnenθ+dnenθ),n=0,1,2,u _ { n } = r ^ { n } \left( c _ { n } e ^ { n \theta } + d _ { n } e ^ { - n \theta } \right) , n = 0,1,2 , \ldots
B) un=rn(cnenθ+dnenθ),n=1,2,3,u _ { n } = r ^ { n } \left( c _ { n } e ^ { n \theta } + d _ { n } e ^ { - n \theta } \right) , n = 1,2,3 , \ldots
C) un=rn(cncos(nθ)+dnsin(nθ)),n=0,1,2,u _ { n } = r ^ { n } \left( c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) \right) , n = 0,1,2 , \ldots
D) un=rncos(nθ),n=0,1,2,u _ { n } = r ^ { n } \cos ( n \theta ) , n = 0,1,2 , \ldots
E) un=rnsin(nθ),n=1,2,3,u _ { n } = r ^ { n } \sin ( n \theta ) , n = 1,2,3 , \ldots
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In the previous three problems, the solution for RR is

A) R=rnR = r ^ { n }
B) R=rn1R = r ^ { - n - 1 }
C) R=c1rn+c2rn1R = c _ { 1 } r ^ { n } + c _ { 2 } r ^ { - n - 1 }
D) R=c1rn+c2rn+1R = c _ { 1 } r ^ { n } + c _ { 2 } r ^ { n + 1 }
E) R=c1rn+c2rn1R = c _ { 1 } r ^ { n } + c _ { 2 } r ^ { n - 1 }
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In the previous problem, the solution of the eigenvalue problem is

A) λ=n2,R=rn,n=0,1,2,\lambda = n ^ { 2 } , R = r ^ { n } , n = 0,1,2 , \ldots
B) λ=n,R=rn,n=0,1,2,\lambda = n , R = r ^ { n } , n = 0,1,2 , \ldots
C) λ=n,Θ=cncos(nθ)+dnsin(nθ),n=0,1,2,.\lambda = n , \Theta = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , .
D) λ=n2,Θ=dnsin(nθ),n=1,2,3,.\lambda = n ^ { 2 } , \Theta = d _ { n } \sin ( n \theta ) , n = 1,2,3 , .
E) λ=n2,Θ=cncos(nθ),n=0,1,2,\lambda = n ^ { 2 } , \Theta = c _ { n } \cos ( n \theta ) , n = 0,1,2 , \ldots
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In separating variables, using, u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) , in the equation 2ur2+2rur+1r22uθ2+cotθr2uθ=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 2 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } + \frac { \cot \theta } { r ^ { 2 } } \frac { \partial u } { \partial \theta } = 0 , the resulting equation for RR and is

A) r2R+2rRλR=0r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda R = 0
B) r2R2rRλR=0r ^ { 2 } R ^ { \prime \prime } - 2 r R ^ { \prime } - \lambda R = 0
C) rR+2rRλrR=0r R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda r R = 0
D) rR+2RλrR=0r R ^ { \prime \prime } + 2 R ^ { \prime } - \lambda r R = 0
E) r2R2rRλrR=0r ^ { 2 } R ^ { \prime \prime } - 2 r R ^ { \prime } - \lambda r R = 0
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In the previous problem, if we also require that Θ\Theta be bounded everywhere, the solution of the eigenvalue problem is

A) λ=n(n+1),Θ=Pn(sinθ),n=1,2,3,\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \sin \theta ) , n = 1,2,3 , \ldots
B) λ=n(n1),Θ=sin(nθ),n=1,2,3,\lambda = n ( n - 1 ) , \Theta = \sin ( n \theta ) , n = 1,2,3 , \ldots
C) λ=n(n1),Θ=cos(nθ),n=1,2,3,\lambda = n ( n - 1 ) , \Theta = \cos ( n \theta ) , n = 1,2,3 , \ldots
D) λ=n(n+1),Θ=Pn(cosθ),n=1,2,3,\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \cos \theta ) , n = 1,2,3 , \ldots
E) λ=n(n1),Θ=Pn(sinθ),n=1,2,3,\lambda = n ( n - 1 ) , \Theta = P _ { n } ( \sin \theta ) , n = 1,2,3 , \ldots
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In changing from Cartesian to polar coordinates, ry\frac { \partial r } { \partial y } is

A) sinθ\sin \theta
B) cosθ\cos \theta
C) sinθ/r\sin \theta / r
D) cosθ/r\cos \theta / r
E) sinθ/r- \sin \theta / r
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In changing from Cartesian to polar coordinates, uy\frac { \partial u } { \partial y } is

A) cosθursinθuθ/r\cos \theta \frac { \partial u } { \partial r } - \sin \theta \frac { \partial u } { \partial \theta } / r
B) cosθur+cosθuθ/r\cos \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta } / r
C) sinθur+cosθuθ\sin \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta }
D) sinθur+cosθuθ/r\sin \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta } / r
E) sinθurcosθuθ/r\sin \theta \frac { \partial u } { \partial r } - \cos \theta \frac { \partial u } { \partial \theta } / r
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In the previous problem, after separating variables using u(r,t)=R(r)T(t)u ( r , t ) = R ( r ) T ( t ) . the resulting problems are

A) rR+R+rλR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + r \lambda R = 0 , R ( 0 ) is bounded R(2)=0,TλT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } - \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
B) rR+R+rλR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + r \lambda R = 0 , R ( 0 ) is bounded R(2)=0,T+λT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
C) rR+RrλR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } - r \lambda R = 0 , R ( 0 ) is bounded R(2)=0,T+λT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
D) rR+R+λR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + \lambda R = 0 , R ( 0 ) is bounded R(2)=0,T+λT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } + \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
E) rR+R+λR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + \lambda R = 0 , R ( 0 ) is bounded R(2)=0,TλT=0,T(0)=0R ( 2 ) = 0 , T ^ { \prime \prime } - \lambda T = 0 , T ^ { \prime } ( 0 ) = 0
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Consider the steady-state temperature distribution in a circular disc of radius CC centere at the origin, with temperature given as a function, f(θ)f ( \theta ) on the boundary r=cr = c and zero on the boundaries θ=0\theta = 0 and θ=π\theta = \pi The mathematical model of this situation is

A) 2ur21rur+1r22uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
B) 2ur2+1rur1r22uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
C) 2ur21rur1r22uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
D) 2ur2+1r2ur+1r2uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
E) 2ur2+1rur+1r22uθ2=0,u(c,θ)=f(θ),u(r,0)=0,u(r,π)=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
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In the previous three problems, the product solutions are

A) u=J0(znr/2)(ancos(znt/2)+bnsin(znt/2))u = J _ { 0 } \left( z _ { n } r / 2 \right) \left( a _ { n } \cos \left( z _ { n } t / 2 \right) + b _ { n } \sin \left( z _ { n } t / 2 \right) \right)
B) u=J0(znr/2)ancos(znt/2)u = J _ { 0 } \left( z _ { n } r / 2 \right) a _ { n } \cos \left( z _ { n } t / 2 \right)
C) u=J0(znr/2)bnsin(znt/2)u = J _ { 0 } \left( z _ { n } r / 2 \right) b _ { n } \sin \left( z _ { n } t / 2 \right)
D) u=J0(nπr)bnsin(nπi)u = J _ { 0 } ( n \pi r ) b _ { n } \sin ( n \pi i )
E) u=J0(nπr)ancos(nπi)u = J _ { 0 } ( n \pi r ) a _ { n } \cos ( n \pi i )
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In the previous problem, the solution of the eigenvalue problem is

A) λ=n,Θ=cncos(nθ)+dnsin(nθ),n=0,1,2,\lambda = n , \Theta = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots
B) λ=n2,=cncos(nθ)+dnsin(nθ),n=0,1,2,\lambda = n ^ { 2 } , \ominus = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots
C) λ=n2,=cncos(nθ)+dnsin(nθ),n=1,2,3,\lambda = n ^ { 2 } , \ominus = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 1,2,3 , \ldots
D) λ=n2,R=rn,n=0,1,2,\lambda = n ^ { 2 } , R = r ^ { n } , n = 0,1,2 , \ldots
E) λ=n,R=rn,n=0,1,2,\lambda = n , R = r ^ { n } , n = 0,1,2 , \ldots
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In changing from Cartesian to polar coordinates, rx\frac { \partial r } { \partial x } is

A) sinθ\sin \theta
B) cosθ\cos \theta
C) sinθ/r\sin \theta / r
D) cosθ/r\cos \theta / r
E) sinθ/r- \sin \theta / r
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In the previous problem, separate variables using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems are

A) r2R+rRλR=0,ΘλΘ=0,Θ(θ)=Θ(θ+2π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } - \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
B) r2R+rR+λR=0,Θ+λΘ=0,Θ(θ)=Θ(θ+2π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
C) r2R+rRλR=0,Θ+λΘ=0,Θ(θ)=Θ(θ+2π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + 2 \pi )
D) r2R+rRλR=0,Θ+λΘ=0,Θ(θ)=Θ(θ+π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } - \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + \pi )
E) r2R+rR+λR=0,Θ+λΘ=0,Θ(θ)=Θ(θ+π)r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , \Theta ^ { \prime \prime } + \lambda \Theta = 0 , \Theta ( \theta ) = \Theta ( \theta + \pi )
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In changing from Cartesian to polar coordinates, ux\frac { \partial u } { \partial x } is

A) sinθur+cosθuθ/r\sin \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta } / r
B) sinθurcosθuθ/r\sin \theta \frac { \partial u } { \partial r } - \cos \theta \frac { \partial u } { \partial \theta } / r
C) cosθursinθuθ/r\cos \theta \frac { \partial u } { \partial r } - \sin \theta \frac { \partial u } { \partial \theta } / r
D) cosθur+cosθuθ/r\cos \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta } / r
E) sinθur+cosθuθ\sin \theta \frac { \partial u } { \partial r } + \cos \theta \frac { \partial u } { \partial \theta }
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The relationships between Cartesian and spherical coordinates are Select all that apply.

A) r2=x2+y2r ^ { 2 } = x ^ { 2 } + y ^ { 2 }
B) tanϕ=y/x\tan \phi = y / x
C) x=rcosϕsinθx = r \cos \phi \sin \theta
D) y=rcosϕcosθy = r \cos \phi \cos \theta
E) z=rcosθz = r \cos \theta
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In the previous three problems, the product solutions are

A) u=J0(nπr/3)cos(nπz/3)u = J _ { 0 } ( n \pi r / 3 ) \cos ( n \pi z / 3 )
B) u=J0(n2π2r/9)cos(nπz/3)u = J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \cos ( n \pi z / 3 )
C) u=J0(n2π2r/9)sin(nπz/3)u = J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \sin ( n \pi z / 3 )
D) u=J0(znr)cosh(zn(z3))u = J _ { 0 } \left( z _ { n } r \right) \cosh \left( z _ { n } ( z - 3 ) \right)
E) u=J0(znr/2)sinh(zn(z3)/2)u = J _ { 0 } \left( z _ { n } r / 2 \right) \sinh \left( z _ { n } ( z - 3 ) / 2 \right)
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Consider the steady-state temperature distribution in a circular cylinder of radius 2 and height 3, with zero temperature at r=2r = 2 and at z=3z = 3 and a temperature of 10 at z=0z = 0 . The mathematical model for this problem is

A) 2ur2+1rur2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
B) 2ur21rur2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
C) 2ur21r2ur+2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
D) 2ur2+1rur+2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
E) 2ur2+1r2ur+2uz2=0,u(2,z)=0,u(r,3)=0,u(r,0)=10\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { \partial ^ { 2 } u } { \partial z ^ { 2 } } = 0 , u ( 2 , z ) = 0 , u ( r , 3 ) = 0 , u ( r , 0 ) = 10
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In the problem 2ur2+2rur+1r22uθ2+cotθr2uθ=0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 2 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } + \frac { \cot \theta } { r ^ { 2 } } \frac { \partial u } { \partial \theta } = 0 , separate variables, using u(r,θ)=R(r)Θ(θ)u ( r , \theta ) = R ( r ) \Theta ( \theta ) . The resulting problems for RR and Θ\Theta are

A) r2R+2rR+λR=0,R(0)r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } + \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
B) r2R+2rRλR=0,R(0)r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
C) r2R+2rRλR=0,R(0)r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θ+cos(θ)Θλsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } - \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
D) r2R2rRλR=0,R(0)r ^ { 2 } R ^ { \prime \prime } - 2 r R ^ { \prime } - \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θ+cos(θ)Θ+λsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } + \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
E) r2R+2rRλR=0,R(0)r ^ { 2 } R ^ { \prime \prime } + 2 r R ^ { \prime } - \lambda R = 0 , R ( 0 ) is bounded; sin(θ)Θcos(θ)Θ+λsin(θ)Θ=0,Θ\sin ( \theta ) \Theta ^ { \prime \prime } - \cos ( \theta ) \Theta ^ { \prime } + \lambda \sin ( \theta ) \Theta = 0,\Theta is bounded on [0,π][ 0 , \pi ] .
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The Laplacian in polar coordinates is

A) 2ur2+1r2ur+1r2uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
B) 2ur2+1rur+1r22uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
C) 2ur21rur+1r22uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
D) 2ur2+1rur1r22uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
E) 2ur21rur1r22uθ2\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } }
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Consider the steady-state temperature distribution in a circular disc of radius CC centere at the origin, with temperature given as a function, f(θ)f ( \theta ) on the boundary. The mathematical model of this situation is

A) 2ur21rur+1r22uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
B) 2ur2+1rur1r22uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
C) 2ur21rur1r22uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
D) 2ur2+1r2ur+1r2uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
E) 2ur2+1rur+1r22uθ2=0,u(c,θ)=f(θ)\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta )
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In the previous four problems, the infinite series solution is (for certain values of the constants)

A) u=n1cnJ0(nπr/3)cos(nπz/3)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } ( n \pi r / 3 ) \cos ( n \pi z / 3 )
B) u=n1cnJ0(n2π2r/9)cos(nπz/3)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \cos ( n \pi z / 3 )
C) u=n1cnJ0(n2π2r/9)sin(nπz/3)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( n ^ { 2 } \pi ^ { 2 } r / 9 \right) \sin ( n \pi z / 3 )
D) u=n1cnJ0(znr/2)sinh(zn(z3)/2)u = \sum _ { n - 1 } ^ { \infty } c _ { n } J _ { 0 } \left( z _ { n } r / 2 \right) \sinh \left( z _ { n } ( z - 3 ) / 2 \right)
E) u=n=1cnJ0(znr)cosh(zn(z3))u = \sum _ { n = 1 } ^ { \infty } c _ { n } J _ { 0 } \left( z _ { n } r \right) \cosh \left( z _ { n } ( z - 3 ) \right)
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In the three previous problems, the product solutions are

A) un=(anrn+bnrn)(cncos(nθ)+dnsin(nθ)),n=0,1,2,u _ { n } = \left( a _ { n } r ^ { n } + b _ { n } r ^ { - n } \right) \left( c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) \right) , n = 0,1,2 , \ldots
B) un=rn(cncos(nθ)+dnsin(nθ)),n=1,2,3,u _ { n } = r ^ { n } \left( c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) \right) , n = 1,2,3 , \ldots
C) un=rn(cncos(nθ)+dnsin(nθ)),n=0,1,2,u _ { n } = r ^ { n } \left( c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) \right) , n = 0,1,2 , \ldots
D) un=rn(cnenθ+dnenθ),n=0,1,2,u _ { n } = r ^ { n } \left( c _ { n } e ^ { n \theta } + d _ { n } e ^ { - n \theta } \right) , n = 0,1,2 , \ldots
E) un=rn(cnenθ+dnenθ),n=1,2,3,u _ { n } = r ^ { n } \left( c _ { n } e ^ { n \theta } + d _ { n } e ^ { - n \theta } \right) , n = 1,2,3 , \ldots
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The solution of r2R+rR+λR=0,R(0)=0,R(1)=0r ^ { 2 } R ^ { \prime \prime } + r R ^ { \prime } + \lambda R = 0 , R ( 0 ) = 0 , R ( 1 ) = 0 is

A) λ=nπ,R=sin(nπlnr)\lambda = n \pi , R = \sin ( n \pi \ln r )
B) λ=(nπ)2,R=sin(nπlnr)\lambda = ( n \pi ) ^ { 2 } , R = \sin ( n \pi \ln r )
C) λ=(nπ)2,R=sin(nπlnr)+cos(nπlnr)\lambda = ( n \pi ) ^ { 2 } , R = \sin ( n \pi \ln r ) + \cos ( n \pi \ln r )
D) λ=nπ,R=sin(nπlnr)cos(nπlnr)\lambda = n \pi , R = \sin ( n \pi \ln r ) - \cos ( n \pi \ln r )
E) none of the above
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In the previous problem, the solution of the eigenvalue problem is

A) λ=n(n+1),R=rn,n=1,2,3,\lambda = n ( n + 1 ) , R = r ^ { n } , n = 1,2,3 , \ldots
B) λ=n(n1),R=rn,n=1,2,3,\lambda = n ( n - 1 ) , R = r ^ { n } , n = 1,2,3 , \ldots
C) λ=n(n1),Θ=Pn(cosθ),n=1,2,3,\lambda = n ( n - 1 ) , \Theta = P _ { n } ( \cos \theta ) , n = 1,2,3 , \ldots
D) λ=n(n+1),Θ=Pn(cosθ),n=1,2,3,\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \cos \theta ) , n = 1,2,3 , \ldots
E) λ=n(n+1),Θ=Pn(sinθ),n=1,2,3,\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \sin \theta ) , n = 1,2,3 , \ldots
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In the previous two problems, the product solutions are

A) un=rnPn(cosθ)u _ { n } = r ^ { n } P _ { n } ( \cos \theta )
B) un=rnPn(cosθ)u _ { n } = r ^ { - n } P _ { n } ( \cos \theta )
C) un=rnPn(sinθ)u _ { n } = r ^ { n } P _ { n } ( \sin \theta )
D) un=rnPn(sinθ)u _ { n } = r ^ { - n } P _ { n } ( \sin \theta )
E) none of the above
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In the previous problem, after separating variables, the resulting problems are

A) rRR+λrR=0,R(0)r R ^ { \prime \prime } - R ^ { \prime } + \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,Z+λZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } + \lambda Z = 0 , Z ( 3 ) = 0
B) rR+RλrR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } - \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,ZλZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } - \lambda Z = 0 , Z ( 3 ) = 0
C) rRRλrR=0,R(0)r R ^ { \prime \prime } - R ^ { \prime } - \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,ZλZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } - \lambda Z = 0 , Z ( 3 ) = 0
D) rR+R+λrR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,ZλZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } - \lambda Z = 0 , Z ( 3 ) = 0
E) rR+R+λrR=0,R(0)r R ^ { \prime \prime } + R ^ { \prime } + \lambda r R = 0 , R ( 0 ) is bounded, R(2)=0,Z+λZ=0,Z(3)=0R ( 2 ) = 0 , Z ^ { \prime \prime } + \lambda Z = 0 , Z ( 3 ) = 0
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When changing from Cartesian to spherical coordinates, ϕx=\frac { \partial \phi } { \partial x } =

A) sinϕ/(rsinθ)- \sin \phi / ( r \sin \theta )
B) sinϕ/(rsinθ)\sin \phi / ( r \sin \theta )
C) cosϕ/(rsinθ)- \cos \phi / ( r \sin \theta )
D) cosϕ/(rcosθ)\cos \phi / ( r \cos \theta )
E) sinϕ/(rcosθ)\sin \phi / ( r \cos \theta )
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In the previous four problems, the infinite series solution of the original problem is u=A0+n=1rn(Ancos(nθ)+Bnsin(nθ))u = A _ { 0 } + \sum _ { n = 1 } ^ { \infty } r ^ { n } \left( A _ { n } \cos ( n \theta ) + B _ { n } \sin ( n \theta ) \right) where Select all that apply.

A) A0=02πf(θ)dθl(2π)A _ { 0 } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) d \theta l ( 2 \pi )
B) An=02πf(θ)sin(nθ)dθ/(cnπ)A _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \sin ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
C) An=02πf(θ)cos(nθ)dθ/(cnπ)A _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \cos ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
D) Bn=02πf(θ)sin(nθ)dθ/(cnπ)B _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \sin ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
E) Bn=02πf(θ)cos(nθ)dθ/(cnπ)B _ { n } = \int _ { 0 } ^ { 2 \pi } f ( \theta ) \cos ( n \theta ) d \theta / \left( c ^ { n } \pi \right)
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39
In the previous problem, the solution of the eigenvalue problem is

A) λ=zn2/4\lambda = z _ { n } ^ { 2 } / 4 , where J0(zn)=0;R=J0(znr/2)J _ { 0 } \left( z _ { n } \right) = 0 ; R = J _ { 0 } \left( z _ { n } r / 2 \right)
B) λ=zn2\lambda = z _ { n } ^ { 2 } , where J0(zn)=0;R=J0(znr)J _ { 0 } \left( z _ { n } \right) = 0 ; R = J _ { 0 } \left( z _ { n } r \right)
C) λ=nπ/9,Z=cos(nπz/3)\lambda = n \pi / 9 , Z = \cos ( n \pi z / 3 )
D) λ=n2π2/9,Z=cos(nπz/3)\lambda = n ^ { 2 } \pi ^ { 2 } / 9 , Z = \cos ( n \pi z / 3 )
E) λ=n2π2/9,Z=sin(nπz/3)\lambda = n ^ { 2 } \pi ^ { 2 } / 9 , Z = \sin ( n \pi z / 3 )
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40
When changing from Cartesian to spherical coordinates, rz=\frac { \partial r } { \partial z } =

A) cosϕ\cos \phi
B) cosθ\cos \theta
C) tanθ\tan \theta
D) sinϕ\sin \phi
E) sinθ\sin \theta
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