Q.49 - Calculus

Question

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

Step-by-Step Solution

Answer

The volume of solids created is

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

Step: 1 Given information

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

Step 2: 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

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

Step 3: 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

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