Use a double integral to find the volume of the solid shown in the figure.

Question

Quizplus · Accepted Answer

Notice that the solid is bounded below by the xy-plane and bounded above by the surface $z=16-x^2-2y^2$. Thus, we can find the volume of the solid using the double integral $$\iint_D \left(16-x^2-2y^2ight)\,dA,$$ where $D$ is the region in the xy-plane bounded by the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$. Switching to polar coordinates, we have \begin{align*} \iint_D \left(16-x^2-2y^2ight)\,dA &= \int_{-\sqrt{2}}^\sqrt{2}\int_{-\sqrt{4-2y^2}}^{\sqrt{4-2y^2}}\left(16-x^2-2y^2ight)\,dx\,dy\ &= \int_0^{2\pi}\int_0^\sqrt{2}\left(16-r^2\cos^2	heta-2r^2\sin^2	hetaight)r\,dr\,d	heta\ &= \int_0^{2\pi}\int_0^\sqrt{2}\left(16r-r^3\cos^2	heta-2r^3\sin^2	hetaight)\,dr\,d	heta\ &= \int_0^{2\pi}\left[\frac{8r^2}{3}-\frac{r^4}{4}\cos^2	heta-\frac{r^4}{2}\sin^2	hetaight]_0^\sqrt{2}\,d	heta\ &= \int_0^{2\pi}\left(\frac{16}{3}-\frac{1}{2}\cos^2	heta-\sin^2	hetaight)\,d	heta\ &= \left[\frac{16	heta}{3}-\frac{	heta}{4}-\frac{\cos	heta\sin	heta}{2}+\frac{	heta}{2}ight]_0^{2\pi}\ &= \frac{32\pi}{3}-\pi\ &= \frac{29}{3}\pi. \end{align*} Therefore, the answer is (C).

Use a Double Integral to Find the Volume of the Solid