The region is bounded above by the unit sphere centered at the origin and below by the plane$z=\frac{3}{5}$

Given a unit sphere centered at the origin and bounded below by the plane $z=\frac{3}{5}$.

The goal of this issue is to discover an iterated integral that depicts the volume of the region circumscribed above by the unit sphere centred at the origin and below by the plane using polar coordinates. $z=3/5$.

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

$z=\frac{3}{5}\text{and}z=\sqrt{1-r^2}$

The equation for the circle of intersection is

$$\begin{gathered} \frac{3}{5}=\sqrt{1-r^2} \\ {r}^2=1-\frac{9}{25} \\ {r}^2=\frac{16}{25} \end{gathered}$$

The radius of the circle of intersection is

$$\begin{gathered} r=\frac{4}{5} \\ \frac{4}{5}\le r\le 1\text{and}0\le \theta \le 2{\pi }^1 \end{gathered}$$

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

$V={\int }_0^{2\pi }{\int }_{4/5}^1\left\{\frac{3}{5}-\sqrt{1-r^2}\right\}rdrd\theta$

Here, $r=\frac{4}{5},r=1$and $\theta =0,\theta =2\pi$

$$\begin{gathered} V={\int }_0^{2\pi }{\int }_{4/5}^1\left\{\frac{3}{5}r-r\sqrt{1-r^2}\right\}drd\theta \\ V={\int }_0^{2\pi }\left[{\int }_{4/5}^1\left\{\frac{3}{5}r-r\sqrt{1-r^2}\right\}dr\right]d\theta \end{gathered}$$

Take the inner integral first.

$$\begin{gathered} {\int }_{4/5}^1\left\{\frac{3}{5}r-r\sqrt{1-r^2}\right\}dr \\ ={\int }_{4/5}^1\frac{3}{5}rdr-{\int }_{45}^{1}r\sqrt{1-r^2}dr \end{gathered}$$

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

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

When $r=4/5$then $t=3/5$and when $r=1$
then $t=0$

Then

$$\begin{gathered} {\int }_{45}^1\frac{3}{5}rdr-{\int }_{4y}^{1}r\sqrt{1-r^2}dr=\frac{3}{5}{\left[\frac{r^2}{2}\right]}_{43}^1-{\int }_{x{s}^0}^{r^{2}dr}\left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right] \\ {\int }_{435}^1\frac{3}{5}rdr-{\int }_{43}^{1}r\sqrt{1-r^2}dr=\frac{3}{5}{\left[\frac{r^2}{2}\right]}_{43}-{\int }_0^{y5}{r}^{2}dt \\ {\int }_{45}^1\frac{3}{5}rdr-{\int }_{4y}^{1}r\sqrt{1-r^2}dr=\frac{3}{5}{\left[\frac{r^2}{2}\right]}_{45}^1-{\left[\frac{r^3}{3}\right]}_0^{3/5} \\ {\int }_{435}^1\frac{3}{5}rdr-{\int }_{43}^{1}r\sqrt{1-r^2}dr=\frac{3}{5}\left[\frac{1}{2}-\frac{16}{50}\right]-\frac{27}{375} \\ {\int }_{43}^1\frac{3}{5}rdr-{\int }_{43}^{1}r\sqrt{1-r^2}dr=\frac{27}{125}-\frac{27}{375} \\ {\int }_{433}^1\frac{3}{5}rdr-{\int }_{43}^{1}r\sqrt{1-r^2}dr=\frac{54}{375} \end{gathered}$$

Therefore,

$$\begin{gathered} V={\int }_0^{2e}\left[\frac{54}{375}\right]d\theta \\ V=\frac{54}{375}[\theta {]}^{2e} \\ V=\frac{54}{375}[2\pi ] \\ V=\frac{108\pi }{375} \end{gathered}$$

the solid's volume is bounded.

$V=\frac{108\pi }{375}$