Solve the problem.
-Let $\mathrm { D }$ be the region bounded below by the $x y$-plane, above by the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 100$, and on the sides by the cylinder $x ^ { 2 } + y ^ { 2 } = 9$. Set up the triple integral in cylindrical coordinates that gives the volume of $D$ using the order of integration $\mathrm { dz } \mathrm {} \mathrm { d } \theta \mathrm { dr }$.
A) $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 9 - r ^ { 2 } } } r d z d \theta d r$
B) $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 9 - r ^ { 2 } } } d z d \theta d r$
C) $\int _ { 0 } ^ { 3 } \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 100 - r ^ { 2 } } } r d z d \theta d r$
D) $\int _ { 0 } ^ { 3 } \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 100 - r ^ { 2 } } } d z d \theta d r$

Question

Quizplus · Accepted Answer

We need to integrate in the order of $dz\ d	heta\ dr$. The integral for cylindrical coordinates in $z$ is $\sqrt{100-r^2}$ from the equation of the sphere: $$z=\sqrt{100-x^2-y^2}=\sqrt{100-r^2}.$$ Note that the inequality $0\leq z\leq \sqrt{100-r^2}$ is satisfied inside $D$. The integral is given by: $$\int_0^{2\pi}\int_0^3\int_0^{\sqrt{100-r^2}} r\ dz\ d	heta\ dr = \int_0^{2\pi}\int_0^3 r\sqrt{100-r^2}\ d	heta\ dr = \frac{9}{2}\int_0^{2\pi}\left[-(100-r^2)^{3/2}ight]_0^3\ d	heta=\frac{9}{2}\int_0^{2\pi}91\ d	heta = 819\pi.$$

Solve the Problem D\mathrm { D }D Be the Region Bounded Below by The

Solve the Problem $\mathrm { D }$ Be the Region Bounded Below by The