The volume of the solid cut out of $z ^ { 2 } + r ^ { 2 } = 16$ by $r = 4 \cos \theta$ is

A) $\frac { 128 ( 3 \pi - 4 ) } { 3 }$
B) $\frac { 128 ( 3 \pi - 4 ) } { 9 }$
C) $\frac { 128 ( 3 \pi + 4 ) } { 9 }$
D) $\frac { 128 ( 3 \pi - 2 ) } { 9 }$
E) $\frac { 128 ( 3 \pi + 2 ) } { 9 }$

Question

The volume of the solid cut out of $z ^ { 2 } + r ^ { 2 } = 16$ by $r = 4 \cos \theta$ is

A) $\frac { 128 ( 3 \pi - 4 ) } { 3 }$
B) $\frac { 128 ( 3 \pi - 4 ) } { 9 }$
C) $\frac { 128 ( 3 \pi + 4 ) } { 9 }$
D) $\frac { 128 ( 3 \pi - 2 ) } { 9 }$
E) $\frac { 128 ( 3 \pi + 2 ) } { 9 }$

Quizplus · Accepted Answer

The volume is calculated by integrating the volume of the solid of revolution formed by the intersection of the cylinder $z^2 + r^2 = 16$ and the circle $r = 4 \cos \theta$. The correct integration results in the volume $\frac { 128 ( 3 \pi - 4 ) } { 9 }$.

The Volume of the Solid Cut Out Of z2+r2=16z ^ { 2 } + r ^ { 2 } = 16z2+r2=16

The Volume of the Solid Cut Out Of $z ^ { 2 } + r ^ { 2 } = 16$