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 $r=3 - x$. For the curves $x = 2 - y^{2}$ and $x = y$, the volume $V$ using the shell method with respect to $y$ is $V=2\pi\int_{a}^{b}r\cdot hdy$, where $a=-2$, $b = 1$, $r=3 - x$ (and we express $x$ in terms of $y$), and $h=(2 - y^{2}-y)$. We rewrite the formula as $V = 2\pi\int_{-2}^{1}(3 - y)(2 - y^{2}-y)dy$. Expand the integrand: $(3 - y)(2 - y^{2}-y)=6-3y^{2}-3y - 2y + y^{3}+y^{2}=y^{3}-2y^{2}-5y + 6$.

Step3: Integrate

$\int(y^{3}-2y^{2}-5y + 6)dy=\frac{y^{4}}{4}-\frac{2y^{3}}{3}-\frac{5y^{2}}{2}+6y+C$.

Step4: Evaluate the definite - integral

$V = 2\pi\left[\frac{y^{4}}{4}-\frac{2y^{3}}{3}-\frac{5y^{2}}{2}+6y\right]_{-2}^{1}$ $=2\pi\left[\left(\frac{1}{4}-\frac{2}{3}-\frac{5}{2}+6\right)-\left(\frac{16}{4}+\frac{16}{3}-\frac{20}{2}-12\right)\right]$ $=2\pi\left[\left(\frac{3 - 8 - 30 + 72}{12}\right)-\left(4+\frac{16}{3}-10 - 12\right)\right]$ $=2\pi\left[\frac{37}{12}-\left(4+\frac{16}{3}-22\right)\right]$ $=2\pi\left[\frac{37}{12}-\left(\frac{12 + 64-264}{12}\right)\right]$ $=2\pi\left[\frac{37-( - 188)}{12}\right]$ $=2\pi\times\frac{225}{12}=\frac{75\pi}{2}\approx117.810$.

Answer:

$117.810$