find the volume of the solid obtained by rotating the region bounded by x = 8 - y² and x = 2y about the line…

Explanation:

Step1: Find intersection points

Set $8 - y^{2}=2y$, then $y^{2}+2y - 8 = 0$. Factoring gives $(y + 4)(y - 2)=0$. So $y=-4$ and $y = 2$.

Step2: Use the method of cylindrical - shells (in terms of y)

The radius of a shell is $r=9 - x$. For the curves $x_1 = 8 - y^{2}$ and $x_2 = 2y$, the volume formula using the shell method $V=\pi\int_{a}^{b}[(r_1)^{2}-(r_2)^{2}]dy$, where $r_1 = 9-(8 - y^{2})=1 + y^{2}$ and $r_2=9 - 2y$, and $a=-4$, $b = 2$.

Step3: Set up the integral

$V=\pi\int_{-4}^{2}[(1 + y^{2})^{2}-(9 - 2y)^{2}]dy=\pi\int_{-4}^{2}(1 + 2y^{2}+y^{4}-(81-36y + 4y^{2}))dy=\pi\int_{-4}^{2}(y^{4}-2y^{2}+36y - 80)dy$.

Step4: Integrate term - by - term

$\int(y^{4}-2y^{2}+36y - 80)dy=\frac{y^{5}}{5}-\frac{2y^{3}}{3}+18y^{2}-80y+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\left(\frac{y^{5}}{5}-\frac{2y^{3}}{3}+18y^{2}-80y\right)\big|_{-4}^{2}\right]$. First, substitute $y = 2$: $\frac{2^{5}}{5}-\frac{2\times2^{3}}{3}+18\times2^{2}-80\times2=\frac{32}{5}-\frac{16}{3}+72 - 160$. Then substitute $y=-4$: $\frac{(-4)^{5}}{5}-\frac{2\times(-4)^{3}}{3}+18\times(-4)^{2}-80\times(-4)=-\frac{1024}{5}+\frac{128}{3}+288 + 320$. $V=\pi\left[\left(\frac{32}{5}-\frac{16}{3}+72 - 160\right)-\left(-\frac{1024}{5}+\frac{128}{3}+288 + 320\right)\right]$. $V=\pi\left(\frac{32 + 1024}{5}-\frac{16+128}{3}+72 - 288-160 - 320\right)$. $V=\pi\left(\frac{1056}{5}-\frac{144}{3}-696\right)$. $V=\pi\left(\frac{1056}{5}-48 - 696\right)=\pi\left(\frac{1056}{5}-744\right)=\pi\left(\frac{1056 - 3720}{5}\right)=\pi\times\frac{-2664}{5}$. $V=\frac{2664\pi}{5}\approx1673.864$.

Answer:

$1673.864$