The region enclosed by the paraboloids $z=x^2+y^2$and $z=16-x^2-y^2$

The given paraboloids $z=x^2+y^2$and$z=16-x^2-y^2$

The goal of this task is to find an iterated integral representing the volume of the solid defined by the paraboloids using polar coordinates.$z=x^2+y^2$and $z=16-x^2-y^2$.

Convert the Cartesian form of paraboloids into polar form.

Substitute $x=r\cos \theta$and $y=r\sin \theta$in the equations of paraboloids

$z=r^2{\cos }^2\theta +r^2{\sin }^2\theta$and $z=16-r^2{\cos }^2\theta -r^2{\sin }^2\theta$

$z=r^2$and $z=16-r^2$

The equation for the circle of intersection is

$$\begin{gathered} x^2+y^2=16-x^2-y^2 \\ {x}^2+y^2=8 \\ {r}^2=8 \\ {r}^2=8 \end{gathered}$$

The radius of the intersecting circle is

$r=2\sqrt{2}$

This implies

$0\le r\le 2\sqrt{2}\text{and}0\le \theta \le 2\pi$

The iterated integral representing the volume can be expressed by the integral of the difference of two given functions.

$V={\int }_a^{2n}{\int }_0^{2\sqrt{2}}\left(\left(16-r^2\right)-r^2\right)rdrd\theta$

Here, $r=0,r=2\sqrt{2}$and $\theta =0,\theta =2\pi$

$V={\int }_a^{2\pi }{\int }_0^{2\sqrt{2}}\left\{16r-2r^3\right\}drd\theta$

Integrate with respect to $r$first.

$$\begin{gathered} V={\int }_a^{2\pi }{\left[\frac{16r^2}{2}-\frac{2r^4}{4}\right]}^{2\sqrt{2}}d\theta \left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right] \\ V={\int }_0^{2n}{\left[8r^2-\frac{r^4}{2}\right]}_0^{2\sqrt{2}}d\theta \end{gathered}$$

$$\begin{gathered} V={\int }_a^{2x}\left[8(2\sqrt{2}{)}^2-\frac{\left.(2\sqrt{2}{)}^4\right]}{2}d\theta \right. \\ V={\int }_0^{2x}[64-32]d\theta \end{gathered}$$

$V={\int }_a^{2x}32d\theta$

Now integrate with respect to $\theta$

$$\begin{gathered} V=32[\theta {]}_3^{2v} \\ V=64\pi \end{gathered}$$

Thus, the volume of solid generated is

$V=64\pi$