Use a double integral in polar coordinates to find the volume of the solid inside the hemisphere $z = \sqrt { 64 - x ^ { 2 } - y ^ { 2 } }$ but outside the cylinder $x ^ { 2 } + y ^ { 2 } = 25$
A) $\frac { 2 } { 3 } 39 ^ { 3 / 2 }$
B) $\frac { 2 } { 3 } 39 ^ { 3 / 2 } \pi$
C) $\frac { 4 } { 3 } 39 ^ { 3 / 2 } \pi$
D) $\frac { 2 } { 3 } 39 ^ { 1 / 2 } \pi$
E) $\frac { 3 } { 2 } 39 ^ { 3 / 2 }$

Question

Quizplus · Accepted Answer

The volume of the solid can be found using a double integral in polar coordinates. The limits for $r$ are from $5$ (the radius of the cylinder) to $8$ (the radius of the hemisphere, since $z = \sqrt{64 - x^2 - y^2}$ implies $x^2 + y^2 + z^2 = 64$, which is a sphere of radius $8$). The limits for $\theta$ are from $0$ to $2\pi$. The integrand is the height of the hemisphere above the xy-plane, which in polar coordinates is $\sqrt{64 - r^2}$. Thus, the volume integral in polar coordinates is $\int_{0}^{2\pi} \int_{5}^{8} r\sqrt{64 - r^2} dr d\theta$. Solving this integral gives the volume of the solid, which is $\frac{2}{3}39^{3/2}\pi$, matching choice B.

Use a Double Integral in Polar Coordinates to Find the Volume