find the volume of the solid obtained by rotating the region bounded by x = 2 - y² and x = y about the line…

Explanation:

Step1: Find intersection points

Set $2 - y^{2}=y$, then $y^{2}+y - 2=0$. Factoring gives $(y + 2)(y - 1)=0$. So $y=-2$ and $y = 1$.

Step2: Use the method of cylindrical - shells

The radius of a shell is $r=6 - x$. For $x = 2 - y^{2}$, the radius is $r_1=6-(2 - y^{2})=4 + y^{2}$, and for $x = y$, the radius is $r_2=6 - y$. The height of the shell $h=(2 - y^{2})-y$. The volume formula using the method of cylindrical - shells for rotating about the line $x = 6$ is $V=\int_{a}^{b}2\pi r h\ dy$, where $a=-2$, $b = 1$. [ \begin{align*} V&=2\pi\int_{-2}^{1}[(6 - y)- (6-(2 - y^{2}))]dy\ &=2\pi\int_{-2}^{1}[(6 - y)-(4 + y^{2})]dy\ &=2\pi\int_{-2}^{1}(2 - y - y^{2})dy \end{align*} ]

Step3: Integrate

[ \begin{align*} \int(2 - y - y^{2})dy&=2y-\frac{y^{2}}{2}-\frac{y^{3}}{3}+C\ \end{align*} ] [ \begin{align*} V&=2\pi\left[\left(2y-\frac{y^{2}}{2}-\frac{y^{3}}{3}\right)\big|_{-2}^{1}\right]\ &=2\pi\left[\left(2\times1-\frac{1^{2}}{2}-\frac{1^{3}}{3}\right)-\left(2\times(-2)-\frac{(-2)^{2}}{2}-\frac{(-2)^{3}}{3}\right)\right]\ &=2\pi\left[\left(2-\frac{1}{2}-\frac{1}{3}\right)-\left(-4 - 2+\frac{8}{3}\right)\right]\ &=2\pi\left[\left(\frac{12 - 3 - 2}{6}\right)-\left(\frac{-12 - 6 + 8}{3}\right)\right]\ &=2\pi\left[\frac{7}{6}-\left(-\frac{10}{3}\right)\right]\ &=2\pi\left[\frac{7 + 20}{6}\right]\ &=2\pi\times\frac{27}{6}\ &=9\pi\approx28.274 \end{align*} ]

Answer:

$28.274$