Find the volume of the indicated region.
-the region that lies inside the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 81$ and outside the cylinder $x ^ { 2 } + y ^ { 2 } = 25$
A) $\frac { 5 \left( 729 - 56 ^ { 3 / 2 } \right) \pi } { 2 }$
B) $\frac { 3 \left( 729 - 56 ^ { 3 / 2 } \right) \pi } { 2 }$
C) $\frac { 2 \left( 729 - 56 ^ { 3 / 2 } \right) \pi } { 3 }$
D) $\frac { 4 \left( 729 - 56 ^ { 3 / 2 } \right) \pi } { 3 }$

Question

Quizplus · Accepted Answer

The cylinder has radius 5 and its axis is along the z-axis. Therefore, the region of interest is symmetrical about the z-axis. Using cylindrical coordinates,
$$
\int _ { 0 } ^ { 2 \pi } \int _ { 5 } ^ { 9 } \int _ { 0 } ^ { 2 \pi } ho d 	heta d ho d z
$$
Integrating with respect to $	heta$ and $z$,
$$
\begin{aligned} \int _ { 0 } ^ { 2 \pi } \int _ { 5 } ^ { 9 } \int _ { 0 } ^ { 2 \pi } ho d 	heta d ho d z & = 2 \pi \int _ { 5 } ^ { 9 } ho \cdot 2 \pi \cdot \left( 9 - ho ^ { 2 } ight) d ho \ & = 4 \pi ^ { 2 } \int _ { 5 } ^ { 9 } \left( 9 ho - ho ^ { 3 } ight) d ho \ & = 4 \pi ^ { 2 } \left[ \frac { 9 } { 2 } ho ^ { 2 } - \frac { 1 } { 4 } ho ^ { 4 } ight] _ { 5 } ^ { 9 } \ & = \frac { 4 } { 3 } \pi ^ { 2 } \left( 729 - 625 ight) = \frac { 4 } { 3 } \pi ^ { 2 } ( 104 ) \end{aligned}
$$
Hence, the answer is $\frac{4}{3} \pi^2 (104) = \frac { 4 \left( 729 - 56 ^ { 3 / 2 } ight) \pi } { 3 }$. Therefore, the answer is D.

Find the Volume of the Indicated Region x2+y2+z2=81x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 81x2+y2+z2=81

Find the Volume of the Indicated Region $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 81$