91影视

The given paraboloids are $z=x^2+y^2$and .$z=R^2-x^2-y^2$

The goal of this issue is to find an iterated integral that expresses the volume of the solid defined by the paraboloids using polar coordinates. $z=x^2+y^2$and$z=R^2-x^2-y^2.$

Convert the Cartesian form of paraboloids into polar form.

Substitute $x=r\cos 胃$and $y=r\sin 胃$in the equations of paraboloids

$z=r^2{\cos }^2胃+r^2{\sin }^2胃$and $z=R^2-r^2{\cos }^2胃-r^2{\sin }^2胃$

$z=r^2$and$z=R^2-r^2$

The equation for the circle of intersection is

$x^2+y^2=R^2-x^2-y^2$

$x^2+y^3=\frac{R^3}{2}$

$r^2=\frac{R^2}{2}$

The radius of the circle of intersection is

$r=\frac{R}{\sqrt{2}}$

This implies

$0鈮�r鈮�\frac{R}{\sqrt{2}}\text{and}0鈮�胃鈮�2蟺$

Therefore, the iterated integral representing the volume can be expressed by the integral of the difference of two given functions.

$V={鈭�}_0^{2蟺}{鈭�}_0^{\frac{蟺}{2}}\left\{\left(R^2-r^2\right)-r^2\right\}rdrd胃$

Here, $r=0,r=\frac{R}{\sqrt{2}}$and $胃=0,胃=2蟺$

$V={鈭�}_0^{2蟺}{鈭�}_0^{N蟺\sqrt{2}}\left\{R^{2}r-2r^3\right\}drd胃$

Integrate with respect to$\mathrm{r}$first.

$$\begin{gathered} V={鈭�}_0^{2蟺}{\left[\frac{R^2{r}^2}{2}-\frac{2r^4}{4}\right]}_0^{e^2\sqrt{2}}d胃\left[鈭�x^{n}dx=\frac{x^{n+1}}{n+1}+C\right] \\ V={鈭�}_0^{2n}\left[\frac{R^4}{4}-\frac{R^4}{8}\right]d胃 \\ V={鈭�}_0^{2蟺}\left[\frac{R^4}{8}\right]d胃 \end{gathered}$$

Now integrate with respect to $胃$

$$\begin{gathered} V=\frac{R^4}{8}[胃{]}_3^{2x} \\ V=\frac{蟺R^4}{4} \end{gathered}$$

Thus, the volume of solid generated is$V=\frac{蟺R^4}{4}$