Question

The region between the cone with an equation $z=\sqrt{x^2+y^2}$and the unit sphere centered at the origin.

Short answer

The volume of solids created is

$V=\left(\frac{1}{2}-\frac{1}{3\sqrt{2}}\right)\pi$

Step-by-step solution

Step: 1 Given information

The given cone with an equation $z=\sqrt{x^2+y^2}$and the unit sphere centered at the origin.

Calculation

The goal of this issue is to find an iterated integral that depicts the volume of the region between the cones using polar coordinates. $z=\sqrt{x^2+y^2}$and the unit sphere is centered at the origin.

The equation of the unit sphere is $x^2+y^2+z^2=1$

Convert the Cartesian form into a polar form.

Substitute $x=r\cos \theta$and $y=r\sin \theta$in the Cartesian forms.

The cone $z=\sqrt{x^2+y^2}$in polar form is $z=r.$

The unit sphere is $x^2+y^2+z^2=1$in polar form is $z=\sqrt{1-r^2}.$

$z-r$and $z=\sqrt{1-r^2}$

The intersecting circle's equation is

$$\begin{gathered} r=\sqrt{1-r^2} \\ 2r^2=1 \\ {r}^2=\frac{1}{2} \end{gathered}$$

The radius of the intersecting circle is

$r=\frac{1}{\sqrt{2}}$

This implies

$0\le r\le \frac{1}{\sqrt{2}}\text{and}0\le \theta \le 2\pi$

Further calculation

The iterated integral expressing volume can be expressed as the integral of the difference between two supplied functions.

$V={\int }_0^{2\pi }{\int }_0^{\frac{1}{\sqrt{2}}}\left\{\sqrt{1-r^2}-r\right\}rdrd\theta$

Here, $r=0,r=\frac{1}{\sqrt{2}}$and $\theta =0,\theta =2\pi$

$$\begin{gathered} V={\int }_0^{2\pi }{\int }_0^{\frac{1}{\sqrt{2}}}\left\{r\sqrt{1-r^2}-r^2\right\}drd\theta \\ V={\int }_0^{2\pi }\left[{\int }_0^{\frac{1}{2}}\left\{r\sqrt{1-r^2}-r^2\right\}dr\right]d\theta \end{gathered}$$

Take the inner integral first.

$$\begin{gathered} {\int }_0^{\frac{1}{\sqrt{2}}}\left\{r\sqrt{1-r^2}-r^2\right\}dr \\ ={\int }_0^{\frac{1}{2}}r\sqrt{1-r^2}dr-{\int }_0^{\frac{1}{2}}{r}^{2}dr \end{gathered}$$

Put simply,$1-r^2=t^2$

$$\begin{gathered} -2rdr=2tdt \\ \Rightarrow rdr=-tdt \end{gathered}$$

When $r=0$then $t=1$and when $r=1/\sqrt{2}$then $t=1/\sqrt{2}$

Then

$$\begin{gathered} {\int }_0^{\frac{1}{\sqrt{2}}}r\sqrt{1-r^2}dr-{\int }_0^{\frac{1}{\sqrt{2}}{r}^2}dr=-{\int }_1^{\frac{1}{2}}tdr-{\left[\frac{r^3}{3}\right]}_0^{1\sqrt{2}}\left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right] \\ {\int }_0^{\frac{1}{2}}r\sqrt{1-r^2}dr-{\int }_0^{\frac{1}{\sqrt{2}}}{r}^{2}dr={\int }_1^{1}ddt-{\left[\frac{r^3}{3}\right]}_0^{1\sqrt{2}}{\int }_0^{\frac{1}{\sqrt{2}}}r\sqrt{1-r^2}dr-{\int }_0^{\frac{1}{\sqrt{2}}{r}^2}dr \\ ={\left[\frac{t^2}{2}\right]}_{U\sqrt{2}}^1-{\left[\frac{r^3}{3}\right]}_0^{2\sqrt{2}}{\int }_0^{\frac{1}{\sqrt{2}}r}r\sqrt{1-r^2}dr-{\int }_0^{\frac{1}{\sqrt{2}}}{r}^{2}dr \\ =\frac{1}{4}-\frac{1}{6\sqrt{2}} \end{gathered}$$

Therefore,

$$\begin{gathered} V={\int }_0^{2\pi }\left[{\int }_0^{\frac{1}{2}}\left\{r\sqrt{1-r^2}-r^2\right\}dr\right]d\theta ={\int }_0^{2x}\left[\frac{1}{4}-\frac{1}{6\sqrt{2}}\right]d\theta \\ V={\left[\frac{\theta }{4}-\frac{\theta }{6\sqrt{2}}\right]}_0^{2\pi } \\ V=\left(\frac{2\pi }{4}-\frac{2\pi }{6\sqrt{2}}\right) \\ V=\left(\frac{1}{2}-\frac{1}{3\sqrt{2}}\right)\pi \end{gathered}$$

Thus, the volume of the solid generated is

$V=\left(\frac{1}{2}-\frac{1}{3\sqrt{2}}\right)\pi$