Question

The region bounded below by the graph of the cone with an equation and bounded above by the plane$z=h$, where $h>0.$

Short answer

The solid's volume is bound.

$V=\frac{\pi h^3}{3}$

Step-by-step solution

Step.1: Given information

The given equation is

$z=\sqrt{x^2+y^2}$

Step.2 Simplification 

The goal of this task is to obtain an iterated integral that depicts the volume of the region confined below the cone and above the plane using polar coordinates. $z=h$

The equation for the cone is

Convert from Cartesian to polar form.

In the Cartesian forms, they substitute$x=r\cos \theta$and $y=r\sin \theta$

$z=h$and $z=r$are the polar forms of the cone.$z=\sqrt{x^2+y^2}$

As, $h>0$, $0\le r\le h$and$0\le \theta \le 2\pi$.are the equations of the circle of intersection.

The integral of the difference between two provided functions can be used to express the iterated integral expressing volume.

$V={\int }_0^{2e}{\int }_0^t\{h-r\}rdrd\theta$

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

$$\begin{gathered} V={\int }_0^{2e}{\int }_0^t\left\{rh-r^2\right\}drd\theta \\ V={\int }_0^{2r}\left[{\int }_0^k\left\{rh-r^2\right\}dr\right]d\theta \end{gathered}$$

Step.3: Further Simplification

First, consider the inner integral.

$$\begin{gathered} V={\int }_0^{2\pi }{\left[\frac{r^2}{2}h-\frac{r^3}{3}\right]}_0^{6}d\theta \left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right] \\ V={\int }_0^{2\pi }\left[\frac{h^3}{2}-\frac{h^3}{3}\right]d\theta \\ V=\frac{h^3}{6}[\theta {]}_0^{2\pi } \\ V=\frac{h^3}{6}[2\pi ] \\ V=\frac{\pi h^3}{3} \end{gathered}$$

The volume of the solid bound is

$V=\frac{\pi h^3}{3}$