Solve the problem.
-Let D be the smaller cap cut from a solid ball of radius 9 units by a plane 3 units from the center of the sphere. Set up the triple integral for the volume of D in spherical coordinates.
A) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \tan ^ { - 1 } ( \sqrt { 72 } / 3 ) } \int _ { \sec \varphi } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { d } \varphi \mathrm { d } \varphi \mathrm { d } \theta$
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \tan ^ { - 1 } ( \sqrt { 72 } / 9 ) } \int _ { 3 \sec \varphi } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { de } \mathrm { d } \varphi \mathrm { d } \theta$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \tan ^ { - 1 } ( \sqrt { 72 } / 9 ) } \int _ { \sec \varphi } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { de } d \varphi d \theta$
D) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \tan ^ { - 1 } ( \sqrt { 72 } / 3 ) } \int _ { 3 \sec \varphi } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { d } \operatorname { d } \varphi \mathrm { d } \theta$

Question

Solve the problem.
-Let D be the smaller cap cut from a solid ball of radius 9 units by a plane 3 units from the center of the sphere. Set up the triple integral for the volume of D in spherical coordinates. 
A) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \tan ^ { - 1 } ( \sqrt { 72 } / 3 ) } \int _ { \sec \varphi } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { d } \varphi \mathrm { d } \varphi \mathrm { d } \theta$
B) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \tan ^ { - 1 } ( \sqrt { 72 } / 9 ) } \int _ { 3 \sec \varphi } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { de } \mathrm { d } \varphi \mathrm { d } \theta$
C) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \tan ^ { - 1 } ( \sqrt { 72 } / 9 ) } \int _ { \sec \varphi } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { de } d \varphi d \theta$
D) $\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \tan ^ { - 1 } ( \sqrt { 72 } / 3 ) } \int _ { 3 \sec \varphi } ^ { 9 } \varrho ^ { 2 } \sin \varphi \mathrm { d } \operatorname { d } \varphi \mathrm { d } \theta$

Quizplus · Accepted Answer

In spherical coordinates, the volume element is $dV=ho^2\sin\phi\,dho\,d\phi\,d	heta$. 
To set up the triple integral for the volume of $D$ in spherical coordinates, we need to find the bounds for $ho$, $\phi$, and $	heta$.

First, we note that the sphere has radius 9 and the plane is 3 units from the center, so the radius of the intersection circle (which is the boundary of $D$) is $\sqrt{9^2-3^2}=6\sqrt{2}$.

In spherical coordinates, the equation of a sphere of radius $R$ is $ho=R$, while the equation of a plane passing through the origin is $ho\cos\phi=d$, where $d$ is the distance from the origin to the plane. 
So, for the given problem, the equation of the sphere is $ho=9$, while the equation of the plane is $ho\cos\phi=3$. Solving for $ho$, we get $ho=\frac{3}{\cos\phi}$.

Since $D$ is the region inside the sphere and above the plane, the bounds for $ho$ are $\frac{3}{\cos\phi}\leqho\leq 9$.

The plane also divides the sphere into two parts. We are interested in the smaller cap, which is above the plane. This cap is symmetric about the $xy$-plane, so we can limit our attention to the upper hemisphere. The angle between the positive $z$-axis and the plane is $\phi=\cos^{-1}(3/9)=\cos^{-1}(1/3)$, so the bounds for $\phi$ are $0\leq\phi\leq\cos^{-1}(1/3)$.

Finally, since the region is symmetric about the $z$-axis, we can integrate over the entire $xy$-plane. So the bounds for $	heta$ are $0\leq	heta\leq 2\pi$.

Putting everything together, we get the integral $\int_{0}^{2\pi}\int_{0}^{\cos^{-1}(1/3)}\int_{3/\cos\phi}^{9}ho^2\sin\phi\,dho\,d\phi\,d	heta$, which matches choice D.

Solve the Problem B) C) D)