Q.50 - Calculus

Question

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

Step-by-Step Solution

Answer

The volume of solids created is

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

Step:1 Given information

Given the equation of the cone  $z=\sqrt{x^2+y^2}$

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 sphere centered at the origin with radius $R,$

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

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

The equation of circle of intersection is

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

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

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

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

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

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

Take the inner integral first.

${\int }_0^{\frac{2}{\sqrt{2}}}\left\{r\sqrt{R^2-r^2}-r^2\right\}dr={\int }_0^{\frac{R}{\sqrt{2}}}r\sqrt{R^2-r^2}dr-{\int }_0^{\frac{R}{\sqrt{2}}}{r}^{2}dr$

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

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

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

Then

$$\begin{gathered} {\int }_0^{\frac{\pi }{\sqrt{2}}}r\sqrt{R^2-r^2}dr-{\int }_0^{\frac{R}{\sqrt{2}}}{r}^{2}dr=-{\int }_2^{\frac{\pi }{\sqrt{2}}}dd-{\left[\frac{r^3}{3}\right]}_0^{N\sqrt{2}}\left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right. \\ {\int }_0^{\frac{\pi }{\sqrt{2}}}r\sqrt{R^2-r^2}dr-{\int }_0^{\frac{\pi }{2}}{r}^{2}dr={\int }_{k=\sqrt{2}}^{g}dt-{\left[\frac{r^3}{3}\right]}_0^{k·\sqrt{2}} \\ {\int }_0^{\sqrt{2}}r\sqrt{R^2-r^2}dr-{\int }_0^{\frac{\pi }{2}}{r}^{2}dr={\left[\frac{r^2}{2}\right]}_{xN\sqrt{2}}^k-{\left[\frac{r^3}{3}\right]}_0^{R/\sqrt{2}} \\ {\int }_0^{\frac{R}{\sqrt{2}}}r\sqrt{R^2-r^2}dr-{\int }_0^{\frac{R}{\sqrt{2}}}{r}^{2}dr=\frac{R^2}{4}-\frac{R^2}{6\sqrt{2}} \end{gathered}$$

Therefore.

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

Thus, the volume of the solid generated is

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