Question

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

Short answer

The region enclosed by Thus, the volume of the solid generated is

$V=\frac{\pi R^4}{4}$

Step-by-step solution

Step: 1 Given information

The given paraboloids are $z=x^2+y^2$and .$z=R^2-x^2-y^2$

Calculation

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

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 $z=R^2-r^2{\cos }^2\theta -r^2{\sin }^2\theta$

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

The equation for the circle of intersection is

$x^2+y^2=R^2-x^2-y^2$

$x^2+y^3=\frac{R^3}{2}$

$r^2=\frac{R^2}{2}$

The radius of the circle of intersection is

$r=\frac{R}{\sqrt{2}}$

This implies

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

Step:3 further calculation

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

$V={\int }_0^{2\pi }{\int }_0^{\frac{\pi }{2}}\left\{\left(R^2-r^2\right)-r^2\right\}rdrd\theta$

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

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

Integrate with respect to$\mathrm{r}$first.

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

Now integrate with respect to $\theta$

$$\begin{gathered} V=\frac{R^4}{8}[\theta {]}_3^{2x} \\ V=\frac{\pi R^4}{4} \end{gathered}$$

Thus, the volume of solid generated is$V=\frac{\pi R^4}{4}$