91影视

The given figure is

A solid of revolution is being formed by rotating the region around y-axis.

The given figure can be shown as below,

The region required to set up the integral should be the region bound by the graph and y-axis.

For the part of region below x-axis, it is clear that the region is bound by the function for vertical line at $x=2\mathrm{and}y-\mathrm{axis},\mathrm{i}.\mathrm{e}.,x=0$. the interval for the y-variable is $[-3,0]$. Hence, it forms a disk of radius x-units.

Thus, the integral formed by rotation this part of region around y-axis is created as $\mathrm{蟺}{鈭�}_{-3}^0{2}^2\mathrm{dy}$.

Now, the part above x-axis is not bounded by y-axis. But it can be expressed as a washer with outer radius being the graph of $x=2$and inner radius is the graph of the function.

The equation of graph of the function $y=f\left(x\right)$can be rewritten as $x=f^{-1}\left(y\right)$. The y-interval for this region is $[0,3]$

Thus, the integral of the volume formed by rotating the upper part of given region is expressed as $\mathrm{蟺}{鈭�}_0^3{2}^2\mathrm{dy}-\mathrm{蟺}{鈭�}_0^3({\mathrm{f}}^{-1}{\left(\mathrm{y}\right))}^2\mathrm{dy}$

Add both the integrals to form the integral for complete volume of solid of revolution for entire region.

$\mathrm{蟺}{鈭�}_{-3}^0{2}^2\mathrm{dy}+\mathrm{蟺}{鈭�}_0^3{2}^2\mathrm{dy}-\mathrm{蟺}{鈭�}_0^3({\mathrm{f}}^{-1}{\left(\mathrm{y}\right))}^2\mathrm{dy}=\mathrm{蟺}{鈭�}_{-3}^0{2}^2\mathrm{dy}+\mathrm{蟺}{鈭�}_0^3(2^2-\left({\mathrm{f}}^{-1}{\left(\mathrm{y}\right))}^2\right)\mathrm{dy}$