91影视

The given paraboloids $z=x^2+y^2$and$z=16-x^2-y^2$

The goal of this task is to find an iterated integral representing the volume of the solid defined by the paraboloids using polar coordinates.$z=x^2+y^2$and $z=16-x^2-y^2$.

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 $z=16-r^2{\cos }^2胃-r^2{\sin }^2胃$

$z=r^2$and $z=16-r^2$

The equation for the circle of intersection is

$$\begin{gathered} x^2+y^2=16-x^2-y^2 \\ {x}^2+y^2=8 \\ {r}^2=8 \\ {r}^2=8 \end{gathered}$$

The radius of the intersecting circle is

$r=2\sqrt{2}$

This implies

$0鈮�r鈮�2\sqrt{2}\text{and}0鈮�胃鈮�2蟺$

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

$V={鈭�}_a^{2n}{鈭�}_0^{2\sqrt{2}}\left(\left(16-r^2\right)-r^2\right)rdrd胃$

Here, $r=0,r=2\sqrt{2}$and $胃=0,胃=2蟺$

$V={鈭�}_a^{2蟺}{鈭�}_0^{2\sqrt{2}}\left\{16r-2r^3\right\}drd胃$

Integrate with respect to $r$first.

$$\begin{gathered} V={鈭�}_a^{2蟺}{\left[\frac{16r^2}{2}-\frac{2r^4}{4}\right]}^{2\sqrt{2}}d胃\left[鈭�x^{n}dx=\frac{x^{n+1}}{n+1}+C\right] \\ V={鈭�}_0^{2n}{\left[8r^2-\frac{r^4}{2}\right]}_0^{2\sqrt{2}}d胃 \end{gathered}$$

$$\begin{gathered} V={鈭�}_a^{2x}\left[8(2\sqrt{2}{)}^2-\frac{\left.(2\sqrt{2}{)}^4\right]}{2}d胃\right. \\ V={鈭�}_0^{2x}[64-32]d胃 \end{gathered}$$

$V={鈭�}_a^{2x}32d胃$

Now integrate with respect to $胃$

$$\begin{gathered} V=32[胃{]}_3^{2v} \\ V=64蟺 \end{gathered}$$

Thus, the volume of solid generated is

$V=64蟺$