91影视

Consider the given information,

$f\left(x\right)=4-x^2$

To find the volume using the shell method, shells of height determined by $f^{-1}\left(y\right)$ are drawn on the y-axis with the average radius of y. This is done because the region bounded by $f\left(x\right)=4-x^2$and the x-axis from $x=0$to $x=2$are rotated around the y-axis.

To find $f^{-1}\left(y\right)$solve $y=4-x^2$ for x gives $x=\sqrt{4-y}$. So $f^{-1}\left(y\right)=\sqrt{4-y}$ Consequently, using the shell method, localid="1661340089898" $\mathrm{Volume}=2蟺{鈭�}_0^4\left(y\right)\sqrt{4-y}dy.$

To solve the integral let $4-y=t^2,$then localid="1661340115427" $dy=-2tdt$and whenlocalid="1661340119661" $y=0,t=2$and when localid="1661340124203" $y=4,t=0$

Consequently.

localid="1661340204023" $\begin{array}{rcl}\mathrm{Volume}& =& 2蟺{鈭�}_2^0\left(4-t^2\right)\sqrt{t^2}(-2tdt)\\ & =& 4蟺{鈭�}_0^2\left(4t^2-t^4\right)dt\mathrm{changing}\mathrm{the}\mathrm{limits}\mathrm{of}\mathrm{integration}\\ & =& 4蟺{\left[4\frac{t^3}{3}-\frac{t^5}{5}\right]}_0^2\mathrm{up}\mathrm{on}\mathrm{integration}\\ & =& 4蟺\left(\left[\frac{32}{3}-\frac{32}{5}\right]-[0-0]\right)\mathrm{simplifying}\end{array}$

Hence, the volume of the solid is $\frac{512}{15}蟺鈮�34.33蟺$