The region bounded above by the unit sphere centered at the origin and bounded below by the plane$z=h$where $0\le h\le 1$.

The area circumscribed on the top by the unit sphere centred at the origin, and on the bottom by the plane$z=h$

The goal of this problem is to find an iterated integral that represents the volume of the region bounded above by the unit sphere centered at the origin and below by the plane $z=h$using polar coordinates.

Equation of unit sphere is $x^2+y^2+z^2=1$

Convert the Cartesian form into 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 $z=\sqrt{1-r^2}$.

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

The equation of circle of intersection is

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

Radius of the circle of intersection is

$r=\sqrt{1-h^2}$

As $0\le h\le 1$therefore, $0\le r\le 1$and $0\le \theta \le 2\pi$.

the integral of the difference of two supplied functions can be used to express the iterated integral expressing volume.

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

Here, $r=0,r=1$and $\theta =0,\theta =2\pi$

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

Take the inner integral first.

${\int }_0^1\left\{hr-r\sqrt{1-r^2}\right\}dr={\int }_0^{1}hrdr-{\int }_0^{1}r\sqrt{1-r^2}dr$

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

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

When $r=0$then $t=1$and when $r=1$then $t=0$

Then

$$\begin{gathered} {\int }_0^{1}hrdr-{\int }_0^{1}r\sqrt{1-r^2}dr=h{\left[\frac{r^2}{2}\right]}_0^1-{\int }_1^0\left(-t^2\right)dt\left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right] \\ {\int }_0^{1}hrdr-{\int }_0^{1}r\sqrt{1-r^2}dr=h{\left[\frac{r^2}{2}\right]}_0^1-{\int }_0^1{t}^{2}dt \\ {\int }_{05}^{1}hrdr-{\int }_0^{1}r\sqrt{1-r^2}dr=h{\left[\frac{r^2}{2}\right]}_0^1-{\left[\frac{t^3}{3}\right]}_0^1 \\ {\int }_0^{1}hrdr-{\int }_0^{1}r\sqrt{1-r^2}dr=h\left[\frac{1}{2}\right]-\frac{1}{3} \\ {\int }_0^{1}hrdr-{\int }_0^{1}r\sqrt{1-r^2}dr=\frac{3h-2}{3} \end{gathered}$$

Therefore,

$$\begin{gathered} V={\int }_0^{2\pi }\left[\frac{3h-2}{3}\right]d\theta \\ V=\frac{3h-2}{3}[\theta {]}_0^{2\pi } \\ V=\frac{3h-2}{3}[2\pi ] \\ V=\frac{6h-4}{3}\pi \end{gathered}$$

Thus, the volume of the solid bounded is

$V=\frac{6h-4}{3}\pi$