Use cylindrical coordinates to evaluate $\int _ { - 2 } ^ { 2 } \int _ { 0 } ^ { \sqrt { 4 - x ^ { 2 } } } \int _ { 0 } ^ { \sqrt { 16 - x ^ { 2 } - y ^ { 2 } } } z d z d y d x$
A) $12$ $\pi$
B) $112$ $\pi$
C) $\frac { 3 } { 2 }$ $\pi$
D) $14$ $\pi$

Question

Quizplus · Accepted Answer

We note that the region is the upper half of a cylinder of radius 2 and height 4. Therefore, in cylindrical coordinates, we have $0\leq	heta\leq2\pi,$ $0\leq r\leq2,$ and $0\leq z\leq\sqrt{16-r^2}.$ Thus, we have \begin{align*}
\int _ { - 2 } ^ { 2 } \int _ { 0 } ^ { \sqrt { 4 - x ^ { 2 } } } \int _ { 0 } ^ { \sqrt { 16 - x ^ { 2 } - y ^ { 2 } } } z d z d y d x&=\int_0^{2\pi}\int_0^2\int_0^{\sqrt{16-r^2}}zr\,dz\,dr\,d	heta \
&=\int_0^{2\pi}\int_0^2\frac{1}{2}(16-r^2)r\,dr\,d	heta \
&=\int_0^{2\pi}\left(8\int_0^2r-\frac{1}{3}r^3ight)\,d	heta \
&=\int_0^{2\pi}\left(32-\frac{32}{3}ight)\,d	heta \
&=\boxed{14\pi}.
\end{align*}

Use Cylindrical Coordinates to Evaluate ∫−22∫04−x2∫016−x2−y2zdzdydx\int _ { - 2 } ^ { 2 } \int _ { 0 } ^ { \sqrt { 4 - x ^ { 2 } } } \int _ { 0 } ^ { \sqrt { 16 - x ^ { 2 } - y ^ { 2 } } } z d z d y d x∫−22​∫04−x2​​∫016−x2−y2​​zdzdydx

Use Cylindrical Coordinates to Evaluate $\int _ { - 2 } ^ { 2 } \int _ { 0 } ^ { \sqrt { 4 - x ^ { 2 } } } \int _ { 0 } ^ { \sqrt { 16 - x ^ { 2 } - y ^ { 2 } } } z d z d y d x$