91影视

The given equation is

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

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$x=r\cos 胃$and $y=r\sin 胃$

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

As, $h>0$, $0鈮�r鈮�h$and$0鈮�胃鈮�2蟺$.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.

$V={鈭�}_0^{2e}{鈭�}_0^t\{h-r\}rdrd胃$

Here, $r=0,r=h$and $胃=0,胃=2蟺$

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

First, consider the inner integral.

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

The volume of the solid bound is

$V=\frac{蟺h^3}{3}$