question find the volume of the solid obtained by rotating the region bounded by x = - 3 + y² and x = - 2y…

Explanation:

Step1: Find intersection points

Set $-3 + y^{2}=-2y$. Rearrange to $y^{2}+2y - 3=0$. Factor: $(y + 3)(y - 1)=0$. So $y=-3$ and $y = 1$.

Step2: Use the disk - washer method (in terms of y)

The outer radius $R(y)=(-8)-(-3 + y^{2})=5 - y^{2}$, the inner radius $r(y)=(-8)-(-2y)=2y - 8$. The volume formula $V=\pi\int_{a}^{b}(R^{2}(y)-r^{2}(y))dy$, where $a=-3$ and $b = 1$. [ \begin{align*} R^{2}(y)&=(5 - y^{2})^{2}=25-10y^{2}+y^{4}\ r^{2}(y)&=(2y - 8)^{2}=4y^{2}-32y + 64\ R^{2}(y)-r^{2}(y)&=25-10y^{2}+y^{4}-(4y^{2}-32y + 64)\ &=y^{4}-14y^{2}+32y - 39 \end{align*} ]

Step3: Calculate the integral

[ \begin{align*} V&=\pi\int_{-3}^{1}(y^{4}-14y^{2}+32y - 39)dy\ &=\pi\left[\frac{y^{5}}{5}-\frac{14y^{3}}{3}+16y^{2}-39y\right]_{-3}^{1}\ &=\pi\left[\left(\frac{1}{5}-\frac{14}{3}+16 - 39\right)-\left(\frac{-243}{5}+\frac{378}{3}+144 + 117\right)\right]\ &=\pi\left[\left(\frac{3 - 70+240 - 585}{15}\right)-\left(\frac{-243 + 630+2160+1755}{15}\right)\right]\ &=\pi\left[\frac{-412}{15}-\frac{4302}{15}\right]\ &=\pi\left[\frac{-412 - 4302}{15}\right]\ &=\pi\left[\frac{-4714}{15}\right]\ &=\frac{4714\pi}{15}\approx987.526 \end{align*} ]

Answer:

$987.526$