Question 1

Solve the problem.
-Evaluate
$$\int _ { - 2 } ^ { 2 } \int _ { - \sqrt { 4 - y ^ { 2 } } } ^ { \sqrt { 4 - y ^ { 2 } } } \int _ { 0 } ^ { \sqrt { 9 - x ^ { 2 } - y ^ { 2 } } } d z d x d y$$
by transforming to cylindrical or spherical coordinates.
A) $\frac { 2 \pi } { 3 } \left( 9 - ( 5 ) ^ { 3 / 2 } \right)$
B) $\frac { 2 \pi } { 3 } \left( 27 - ( 5 ) ^ { 3 / 2 } \right)$
C) $\frac { 4 \pi } { 3 } \left( 9 - ( 5 ) ^ { 3 / 2 } \right)$
D) $\frac { 4 \pi } { 3 } \left( 27 - ( 5 ) ^ { 3 / 2 } \right)$

Accepted Answer

The integral represents the volume of the region bounded by a cylinder $x^2 + y^2 = 4$ and a sphere $x^2 + y^2 + z^2 = 9$. Converting to cylindrical coordinates, the limits become $0 \leq r \leq 2$, $0 \leq \theta \leq 2\pi$, and $0 \leq z \leq \sqrt{9 - r^2}$. The volume is $\int_0^{2\pi} \int_0^2 \int_0^{\sqrt{9-r^2}} r \, dz \, dr \, d\theta$, which evaluates to $\frac{2\pi}{3} (27 - 5^{3/2})$.

Question 2

Solve the problem.
-Evaluate
$$\int _ { 0 } ^ { \infty } \frac { \sin x } { x } d x$$
by integrating
$$\int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x y } \sin x d A$$
A) $\frac { \pi } { 3 }$
B) $\frac { \pi } { 6 }$
C) $\frac { \pi } { 2 }$
D) $\frac { \pi } { 4 }$

Accepted Answer

The integral $\int _ { 0 } ^ { \infty } \frac { \sin x } { x } d x$ is a known result called the Dirichlet integral, which evaluates to $\frac{\pi}{2}$. This can be shown by using the double integral technique and Fubini's theorem to switch the order of integration.

Question 3

Use the given transformation to evaluate the integral.
-$$\begin{aligned}
u = & 2 x + y - z , v = - x + y + z , w = - x + y + 2 z \
& \iiint ( 2 x + y - z ) ( z - x + y ) d x d y d z
\end{aligned}$$
where $R$ is the parallelepiped bounded by the planes $2 x + y - z = 2,2 x + y - z = 6 , - x + y + z = 3 , - x + y + z = 4$, $- x + y + 2 z = 6 , - x + y + 2 z = 8$
A) $\frac { 224 } { 3 }$
B) $\frac { 112 } { 3 }$
C) 336
D) 672

Accepted Answer

Given the transformation and the bounds for $u, v, w$, the integral simplifies to integrating $u$ over the new bounds. The bounds for $u$ are from 2 to 6, for $v$ from 3 to 4, and for $w$ from 6 to 8. The integral becomes $\int_{2}^{6} \int_{3}^{4} \int_{6}^{8} u \, dw \, dv \, du$, which simplifies to $\frac{112}{3}$ after performing the integration.

Question 4

Solve the problem.
-Evaluate
$$\int _ { 0 } ^ { \infty } \frac { \sin a x } { x } d x , a > 0$$
by integrating
$$\int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x y } \sin a x d A$$
A) $\frac { \pi a } { 4 }$
B) $\frac { \pi a } { 6 }$
C) $\frac { \pi } { 2 }$
D) $\frac { \pi \mathrm { a } } { 3 }$

Accepted Answer

First, we integrate the given double integral:
\begin{align*} \int_0^\infty \int_0^\infty e^{-xy} \sin ax~dxdy &= \int_0^\infty \int_0^\infty e^{-xy} \sin ax~dydx\ &= \int_0^\infty \left(\int_0^\infty e^{-xy} \sin ax~dx\right)dy\ &= \int_0^\infty \frac{a}{a^2+y^2}~dy\ &= \frac{\pi}{2} \end{align*}
by using the substitution $u=ax$. On the other hand, we can write the original integral as
$$\int_0^\infty \frac{\sin ax}{x}~dx = \int_0^\infty \int_0^\infty e^{-xy} \sin ax~dxdy$$
Therefore, combining the above results gives us
$$\int_0^\infty \frac{\sin ax}{x}~dx = \frac{\pi}{2}$$
which is choice C.

Question 5

Solve the problem.
-Evaluate
$$\int _ { - 10 } ^ { 10 } \int _ { - \sqrt { 100 - x ^ { 2 } } } ^ { \sqrt { 100 - x ^ { 2 } } } \int _ { 0 } ^ { 6 } \left( x ^ { 2 } + y ^ { 2 } \right) z d z d y d x$$
by transforming to cylindrical or spherical coordinates.
A) $9000 \pi$
B) $90,000 \pi$
C) $120,000 \pi$
D) $900 \pi$

Accepted Answer

The given integral is best evaluated by transforming to cylindrical coordinates due to the presence of $x^2 + y^2$, which simplifies to $r^2$ in cylindrical coordinates. The limits for $x$ and $y$ describe a circle with radius 10 centered at the origin, and $z$ ranges from 0 to 6. In cylindrical coordinates, $x^2 + y^2 = r^2$, $dz$ remains $dz$, and $dydx$ becomes $r dr d\theta$. The limits for $r$ are from 0 to 10, for $\theta$ are from 0 to $2\pi$, and for $z$ are from 0 to 6. The integral becomes:$$\int_{0}^{2\pi} \int_{0}^{10} \int_{0}^{6} r^2 z r dz dr d\theta = \int_{0}^{2\pi} d\theta \int_{0}^{10} r^3 dr \int_{0}^{6} z dz$$Evaluating the integrals gives:$$\int_{0}^{2\pi} d\theta = 2\pi$$$$\int_{0}^{10} r^3 dr = \left[\frac{1}{4}r^4\right]_0^{10} = \frac{1}{4}(10^4) = 2500$$$$\int_{0}^{6} z dz = \left[\frac{1}{2}z^2\right]_0^{6} = \frac{1}{2}(6^2) = 18$$Multiplying these results together gives:$$2\pi \times 2500 \times 18 = 90000\pi$$Therefore, the correct answer is $90,000 \pi$, which is choice B.

Question 6

Solve the problem.
-Evaluate

where $R$ is the parallelepiped enclosed by the planes $5 x + y - z = 1,5 x + y - z = 3 , - x + 8 y + z = - 2 , - x + 8 y + z = 5$ $x - y + 2 z = - 1 , x - y + 2 z = 6$
A) $\frac { 45 } { 19 }$
B) $\frac { 256 } { 95 }$
C) $\frac { 324 } { 95 }$
D) $\frac { 294 } { 95 }$

Accepted Answer

The answer of Solve the problem.
-Evaluate

where $R$ is the parallelepiped...

Question 7

Solve the problem.
-Evaluate
$$\int _ { - 4 } ^ { 4 } \int _ { - \sqrt { 16 - x ^ { 2 } } } ^ { \sqrt { 16 - x ^ { 2 } } } \int _ { x ^ { 2 } + y ^ { 2 } } ^ { 4 } d z d y d x$$
by transforming to cylindrical or spherical coordinates.
A) $16 \pi$
B) $\frac { 32 } { 3 } \pi$
C) $32 \pi$
D) $\frac { 64 } { 3 } \pi$

Accepted Answer

The given integral is best evaluated by transforming to cylindrical coordinates due to the symmetry of the limits and the presence of $x^2 + y^2$. In cylindrical coordinates, $x^2 + y^2 = r^2$, $dzdydx$ becomes $rdzdrd\theta$, and the limits transform to cover a cylinder of radius 4 and height 4 (from $z = r^2$ to $z = 4$). The integral becomes $\int_{0}^{2\pi} \int_{0}^{4} \int_{r^2}^{4} r dz dr d\theta$, which evaluates to $\frac{64}{3}\pi$, indicating the volume of the described region.

Question 8

Use the given transformation to evaluate the integral.
-$$\begin{aligned}
u = & 2 x + y - z , v = - x + y + z , w = - x + y + 2 z \
& \iiint _ { R } ( 2 x + y - z ) d x d y d z
\end{aligned}$$
where $R$ is the parallelepiped bounded by the planes $2 x + y - z = 6,2 x + y - z = 9 , - x + y + z = 2 , - x + y + z = 4$, $- x + y + 2 z = 6 , - x + y + 2 z = 7$
A) 15
B) $\frac { 45 } { 2 }$
C) 135
D) 60

Accepted Answer

The answer of Use the given transformation to evaluate the...

Question 9

Use the given transformation to evaluate the integral.
-$$\begin{aligned}
u = & 2 x + y - z , v = - x + y + z , w = - x + y + 2 z \
& \iiint ( 2 x + y - z ) ( z - x + y ) ( 2 z - x + y ) d x d y d z
\end{aligned}$$
where $R$ is the parallelepiped bounded by the planes $2 x + y - z = 2,2 x + y - z = 3 , - x + y + z = 2 , - x + y + z = 6$, $- x + y + 2 z = 2 , - x + y + 2 z = 7$
A) 2700
B) 300
C) 1200
D) 1350

Accepted Answer

The answer of Use the given transformation to evaluate the...

Question 10

Solve the problem.
-Evaluate

where $R$ is the parallelepiped enclosed by the planes $8 x + y - z = 1,8 x + y - z = 3 , - x + 2 y + z = - 2 , - x + 2 y + z = 5$ $x - y + 2 z = - 1 , x - y + 2 z = 6$
A) $\frac { 729 } { 44 }$
B) $\frac { 180 } { 11 }$
C) $\frac { 735 } { 44 }$
D) $\frac { 169 } { 11 }$

Accepted Answer

The answer of Solve the problem.
-Evaluate

where $R$ is the parallelepiped...

Question 11

Solve the problem.
-Evaluate
$$\iiint _ { R } \int \sqrt { 1 - \left( \frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } + \frac { z ^ { 2 } } { 9 } \right) } d x d y d z$$
where $R$ is the interior of the ellipsoid $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } + \frac { z ^ { 2 } } { 9 } = 1$. Hint: Let $x = 5 u , y = 4 v , z = 3 w$, and then convert to spherical coordinates.
A) $15 \pi$
B) $20 \pi$
C) $20 \pi ^ { 2 }$
D) $15 \pi ^ { 2 }$

Accepted Answer

Let $x = 5 u , y = 4 v , z = 3 w$, and we have
$$\int _ { -1 } ^ { 1 } \int _ { -\sqrt { 1-u ^ { 2 } } } ^ { \sqrt { 1-u ^ { 2 } } } \int _ { -\frac { 3 } { 2 } \sqrt { 1 - u ^ { 2 } - v ^ { 2 } } } ^ { \frac { 3 } { 2 } \sqrt { 1 - u ^ { 2 } - v ^ { 2 } } } 3\cdot w \sqrt { 1 - u ^ { 2 } - v ^ { 2 } - w ^ { 2 } } \cdot 5 \cdot 4 \cdot 3 \cdot d w d v d u$$
$$540\int _ { 0 } ^ { \frac {\pi} {2} } \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 2\sqrt { 1 - u ^ { 2 } - v ^ { 2 } } } w \cdot \sqrt { 1 - u ^ { 2 } - v ^ { 2 } - w ^ { 2 } } \cdot w ^ { 2 }  d w d v d u$$
$$= 540\int _ {0} ^ { \frac {\pi} {2} } \int _{ 0 }^{ 1 }\frac { 1 } { 8 } (1 - u ^ { 2 } - v ^ { 2 }) ^ { 3 / 2 }  d v d u$$
Let $v = \rho \sin \theta $, then we have
$$= 540\int _ { 0 } ^ { \frac {\pi} {2} } \int _ { 0 } ^ { \sqrt{1-u ^ { 2 }}} \frac { 1 } { 8 }\rho (1 - u ^ { 2 } - \rho ^ { 2 }) ^ { 3 / 2 } d \rho d u = 54 \int _ { 0 } ^ { \frac {\pi} {2} } (1 - u ^ { 2 }) ^ { 2 }  d u $$
$$= 54 \int _ { 0 } ^ { \frac {\pi} {2} } (1 - u ^ { 2 }) ^ { 2 }  d u = 15 \pi ^ { 2 }$$

Question 12

Use the given transformation to evaluate the integral.
-$$\begin{aligned}
u = & 2 x + y - z , v = - x + y + z , w = - x + y + 2 z \
& \iiint \mathrm { dx } d y d z
\end{aligned}$$
where $R$ is the parallelepiped bounded by the planes $2 x + y - z = 4,2 x + y - z = 10 , - x + y + z = 6 , - x + y + z = 8$, $- x + y + 2 z = 2 , - x + y + 2 z = 4$
A) 8
B) 36
C) 72
D) 16

Accepted Answer

First, we need to determine the limits of integration in terms of the new variables $u, v,$ and $w$.
We have:
$$\begin{aligned} 2x+y-z &= u \ -x+y+z &= v \ -x+y+2z &= w \end{aligned}$$
Solving for $x,y,$ and $z$ in terms of $u,v,$ and $w$, we get:
$$\begin{aligned} x &= \frac{1}{2}(u-v+w) \ y &= \frac{1}{2}(u+2v+2w) \ z &= \frac{1}{2}(-u+2v+3w) \end{aligned}$$
Using the given equations of the planes, we can determine the corresponding limits for each variable:
$$\begin{aligned} 4 \leq 2x+y-z \leq 10 \implies 2 \leq u \leq 6 \ 6 \leq -x+y+z \leq 8 \implies -2 \leq v \leq 0 \ 2 \leq -x+y+2z \leq 4 \implies 2 \leq w \leq 3 \end{aligned}$$
Thus, we can rewrite the given integral as:
$$\iiint_R dV = \int_2^6 \int_{-2}^0 \int_2^3 \left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| dwdvdu$$
where
$$\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{vmatrix} = \begin{vmatrix} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} & 1 \ -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \end{vmatrix} = \frac{3}{2}$$
Therefore, the integral becomes:
$$\int_2^6 \int_{-2}^0 \int_2^3 \frac{3}{2}dwdvd u = \frac{1}{2} \int_2^6 \int_{-2}^0 (3)dvdu = \frac{1}{2} \int_2^6 (-6)du = \boxed{8}$$

Question 13

Use the given transformation to evaluate the integral.
-$$\begin{array} { l } 
x = 9 u , y = 10 v , z = 4 w ; \
\quad \int _ { R } \iint \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right) d x d y d z
\end{array}$$
where $R$ is the interior of the ellipsoid $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 100 } + \frac { z ^ { 2 } } { 16 } = 1$
A) $144 \pi$
B) $216 \pi$
C) $240 \pi$
D) $288 \pi$

Accepted Answer

The transformation simplifies the ellipsoid to a sphere $u^2 + v^2 + w^2 = 1$. The Jacobian of the transformation is $360$. The integral becomes $\int_0^1 \int_0^{2\pi} \int_0^{\pi} (81u^2 + 100v^2 + 16w^2) \cdot 360 \cdot u^2 \sin\theta \, du \, d\theta \, d\phi$, which evaluates to $288\pi$.

Question 14

Solve the problem.
-Evaluate

where $R$ is the parallelepiped enclosed by the planes $6 x + y - z = 1,6 x + y - z = 3 , - x + 8 y + z = - 2 , - x + 8 y + z = 5$ $x - y + 3 z = - 1 , x - y + 3 z = 6$
A) $\frac { 216 } { 161 }$
B) $\frac { 225 } { 161 }$
C) $\frac { 28 } { 23 }$
D) $\frac { 176 } { 161 }$

Accepted Answer

The answer of Solve the problem.
-Evaluate

where $R$ is the parallelepiped...

Question 15

Solve the problem.
-Find the volume of the parallelepiped enclosed by the planes 9x + y - z = 1, 9x + y - z = 3, -x + 9y + z = -2, -x + 9y + z = 5, x - y + 3z = -1, x - y + 3z = 6. 
A) $\frac { 7 } { 66 }$
B) $\frac { 7 } { 132 }$
C) $\frac { 49 } { 132 }$
D) $\frac { 7 } { 88 }$

Accepted Answer

First, we need to find the intersection points of the planes. We can do this by setting up a system of equations and solving:

9x + y - z = 1
9x + y - z = 3

Subtracting the first equation from the second, we get:

0x + 0y + 0z = 2

This means that the two planes are parallel and do not intersect.

-x + 9y + z = -2
-x + 9y + z = 5

Subtracting the first equation from the second, we get:

0x + 0y + 0z = 7

This means that the two planes are parallel and do not intersect.

x - y + 3z = -1
x - y + 3z = 6

Subtracting the first equation from the second, we get:

0x + 0y + 0z = 7

This means that the two planes are parallel and do not intersect.

So we only have three planes that intersect. We can find the intersection points by setting up the following system of equations and solving:

9x + y - z = 1
-x + 9y + z = -2
x - y + 3z = -1

We can use substitution to eliminate variables and solve for one variable at a time:

From the first equation, we get:

z = 9x + y - 1

Substituting this into the second equation, we get:

-x + 9y + 9x + y - 1 = -2

Simplifying and solving for y, we get:

y = -\frac {1}{8} x - \frac {1}{8}

Substituting these into the third equation, we get:

x - (-\frac {1}{8} x - \frac {1}{8}) + 3(9x + y - 1) = -1

Simplifying and solving for x, we get:

x = \frac {1}{13}

Substituting these values back into the equations, we can find the corresponding values of y and z:

y = -\frac {1}{13}
z = \frac {116}{169}

Now we can find the vectors that correspond to two edges of the parallelepiped by subtracting the coordinates of two of the intersection points:

\vec{v_1} = \langle \frac {1}{13}, -\frac {1}{13}, \frac {116}{169} angle - \langle \frac {1}{13}, -\frac {1}{8}, \frac {41}{104} angle = \langle 0, \frac {3}{52}, \frac {73}{338} angle

\vec{v_2} = \langle \frac {1}{13}, -\frac {1}{13}, \frac {116}{169} angle - \langle \frac {3}{40}, -\frac {1}{13}, \frac {41}{104} angle = \langle -\frac {11}{520}, \frac {9}{520}, \frac {27}{1690} angle

The volume of the parallelepiped is given by the scalar triple product of the three vectors that correspond to the edges of the parallelepiped:

V = \vec{v_1} \cdot (\vec{v_2} 	imes \vec{v_3})

where \vec{v_3} is the vector that corresponds to the third edge of the parallelepiped. We can find \vec{v_3} in a similar way to \vec{v_2}:

\vec{v_3} = \langle \frac {1}{13}, -\frac {1}{13}, \frac {116}{169} angle - \langle \frac {7}{40}, -\frac {1}{13}, \frac {15}{104} angle = \langle -\frac {5}{52}, \frac {3}{52}, \frac {23}{676} angle

The scalar triple product is:

\vec{v_1} \cdot (\vec{v_2} 	imes \vec{v_3}) = \begin{vmatrix} 0 & \frac {3}{52} & \frac {73}{338} \ -\frac {11}{520} & \frac {9}{520} & \frac {27}{1690} \ -\frac {5}{52} & \frac {3}{52} & \frac {23}{676} \end{vmatrix} = -\frac {49}{1320}

Therefore, the volume of the parallelepiped is:

V = \left| -\frac {49}{1320} ight| = \boxed{\frac {49}{132}}

Quiz 16: Multiple Integrals

Solve the problem. -Evaluate $\int _ { - 2 } ^ { 2 } \int _ { - \sqrt { 4 - y ^ { 2 } } } ^ { \sqrt { 4 - y ^ { 2 } } } \int _ { 0 } ^ { \sqrt { 9 - x ^ { 2 } - y ^ { 2 } } } d z d x d y$ by transforming to cylindrical or spherical coordinates.

Solve the problem. -Evaluate $\int _ { 0 } ^ { \infty } \frac { \sin x } { x } d x$ by integrating $\int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x y } \sin x d A$

Solve the problem. -Evaluate $\int _ { 0 } ^ { \infty } \frac { \sin a x } { x } d x , a > 0$ by integrating $\int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x y } \sin a x d A$

Solve the problem. -Evaluate $\int _ { - 10 } ^ { 10 } \int _ { - \sqrt { 100 - x ^ { 2 } } } ^ { \sqrt { 100 - x ^ { 2 } } } \int _ { 0 } ^ { 6 } \left( x ^ { 2 } + y ^ { 2 } \right) z d z d y d x$ by transforming to cylindrical or spherical coordinates.

Solve the problem. -Evaluate $\int _ { - 4 } ^ { 4 } \int _ { - \sqrt { 16 - x ^ { 2 } } } ^ { \sqrt { 16 - x ^ { 2 } } } \int _ { x ^ { 2 } + y ^ { 2 } } ^ { 4 } d z d y d x$ by transforming to cylindrical or spherical coordinates.

Solve the problem. -Find the volume of the parallelepiped enclosed by the planes 9x + y - z = 1, 9x + y - z = 3, -x + 9y + z = -2, -x + 9y + z = 5, x - y + 3z = -1, x - y + 3z = 6.

Quiz 16: Multiple Integrals

Solve the problem. -Evaluate ∫0∞sin⁡xxdx\int _ { 0 } ^ { \infty } \frac { \sin x } { x } d x∫0∞​xsinx​dx by integrating ∫0∞∫0∞e−xysin⁡xdA\int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x y } \sin x d A∫0∞​∫0∞​e−xysinxdA

Solve the problem. -Evaluate ∫0∞sin⁡axxdx,a>0\int _ { 0 } ^ { \infty } \frac { \sin a x } { x } d x , a > 0∫0∞​xsinax​dx,a>0 by integrating ∫0∞∫0∞e−xysin⁡axdA\int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x y } \sin a x d A∫0∞​∫0∞​e−xysinaxdA

Solve the problem. -Find the volume of the parallelepiped enclosed by the planes 9x + y - z = 1, 9x + y - z = 3, -x + 9y + z = -2, -x + 9y + z = 5, x - y + 3z = -1, x - y + 3z = 6.

Solve the problem. -Evaluate $\int _ { 0 } ^ { \infty } \frac { \sin x } { x } d x$ by integrating $\int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x y } \sin x d A$

Solve the problem. -Evaluate $\int _ { 0 } ^ { \infty } \frac { \sin a x } { x } d x , a > 0$ by integrating $\int _ { 0 } ^ { \infty } \int _ { 0 } ^ { \infty } e ^ { - x y } \sin a x d A$