MATH SOLVE

8 months ago

Q:
# Se polar coordinates to find the volume of the given solid. the solid above the xy-plane that lies below the ellipsoid 4x2 + 4y2 + z2 = 64 but inside the cylinder x2 + y2 = 4

Accepted Solution

A:

[tex]\begin{cases}x=\frac12r\cos t\\\\\frac12r\sin t\end{cases}\implies 4x^2+4y^2+z^2=64\iff z^2=64-r^2[/tex]

Call the region [tex]D[/tex]. Then

[tex]\displaystyle\iiint_D\mathrm dV=\int_{t=0}^{t=2\pi}\int_{r=0}^{r=2}\int_{z=-\sqrt{64-r^2}}^{z=\sqrt{64-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]

[tex]=\displaystyle4\pi\int_{r=0}^{r=2}r\sqrt{64-r^2}\,\mathrm dr[/tex]

[tex]=\displaystyle-2\pi\int_{r=0}^{r=2}\sqrt{64-r^2}(-2r\,\mathrm dr)[/tex]

[tex]=\displaystyle-2\pi\int_{r=0}^{r=2}\sqrt{64-r^2}\,\mathrm d(64-r^2)=\left(\dfrac{2048}3-16\sqrt{15}\right)\pi\approx197.887[/tex]

Call the region [tex]D[/tex]. Then

[tex]\displaystyle\iiint_D\mathrm dV=\int_{t=0}^{t=2\pi}\int_{r=0}^{r=2}\int_{z=-\sqrt{64-r^2}}^{z=\sqrt{64-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]

[tex]=\displaystyle4\pi\int_{r=0}^{r=2}r\sqrt{64-r^2}\,\mathrm dr[/tex]

[tex]=\displaystyle-2\pi\int_{r=0}^{r=2}\sqrt{64-r^2}(-2r\,\mathrm dr)[/tex]

[tex]=\displaystyle-2\pi\int_{r=0}^{r=2}\sqrt{64-r^2}\,\mathrm d(64-r^2)=\left(\dfrac{2048}3-16\sqrt{15}\right)\pi\approx197.887[/tex]