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 $2 - y^{2}=-y$. Rearrange to $y^{2}-y - 2 = 0$. Factor: $(y - 2)(y+1)=0$. So $y=-1$ and $y = 2$.

Step2: Use the method of cylindrical - shells (or the washer method in terms of $y$).

The outer radius $R(y)=9-(-y)=9 + y$ and the inner radius $r(y)=9-(2 - y^{2})=7 + y^{2}$ for $y\in[-1,2]$. The volume $V=\pi\int_{-1}^{2}[(R(y))^{2}-(r(y))^{2}]dy=\pi\int_{-1}^{2}[(9 + y)^{2}-(7 + y^{2})^{2}]dy$. Expand the integrand: $(9 + y)^{2}=81+18y + y^{2}$ and $(7 + y^{2})^{2}=49 + 14y^{2}+y^{4}$. So $(9 + y)^{2}-(7 + y^{2})^{2}=81+18y + y^{2}-(49 + 14y^{2}+y^{4})=-y^{4}-13y^{2}+18y + 32$.

Step3: Integrate

$\int(-y^{4}-13y^{2}+18y + 32)dy=-\frac{1}{5}y^{5}-\frac{13}{3}y^{3}+9y^{2}+32y+C$. Evaluate the definite - integral: $\left(-\frac{1}{5}y^{5}-\frac{13}{3}y^{3}+9y^{2}+32y\right)\big|_{-1}^{2}$ $=\left(-\frac{1}{5}(2)^{5}-\frac{13}{3}(2)^{3}+9(2)^{2}+32(2)\right)-\left(-\frac{1}{5}(-1)^{5}-\frac{13}{3}(-1)^{3}+9(-1)^{2}+32(-1)\right)$ $=\left(-\frac{32}{5}-\frac{104}{3}+36 + 64\right)-\left(\frac{1}{5}+\frac{13}{3}+9-32\right)$ $=\left(-\frac{32}{5}-\frac{104}{3}+100\right)-\left(\frac{1}{5}+\frac{13}{3}-23\right)$ $=-\frac{32}{5}-\frac{104}{3}+100-\frac{1}{5}-\frac{13}{3}+23$ $=\left(-\frac{32}{5}-\frac{1}{5}\right)+\left(-\frac{104}{3}-\frac{13}{3}\right)+(100 + 23)$ $=-\frac{33}{5}-\frac{117}{3}+123$ $=-\frac{33}{5}-39 + 123$ $=-\frac{33}{5}+84$ $=\frac{-33 + 420}{5}=\frac{387}{5}$. Then $V=\pi\times\frac{387}{5}\approx\frac{3.14159\times387}{5}\approx243.163$.

Answer:

$243.163$