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

Explanation:

Step1: Find intersection points

Set $-3 + y^{2}=-2y$, which can be rewritten as $y^{2}+2y - 3=0$. Factoring gives $(y + 3)(y - 1)=0$. So $y=-3$ and $y = 1$.

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

The outer - radius $R(y)$ and inner - radius $r(y)$ for rotation about $x=-7$. The distance from a point $x$ to the line $x = - 7$ is $d=x + 7$. So $R(y)=(-2y+7)$ and $r(y)=(-3 + y^{2}+7)=y^{2}+4$.

Step3: Apply the volume formula

The volume $V$ of the solid of revolution using the washer method $V=\pi\int_{a}^{b}(R^{2}(y)-r^{2}(y))dy$, where $a=-3$, $b = 1$. [ \begin{align*} V&=\pi\int_{-3}^{1}[(-2y + 7)^{2}-(y^{2}+4)^{2}]dy\ &=\pi\int_{-3}^{1}(4y^{2}-28y + 49-(y^{4}+8y^{2}+16))dy\ &=\pi\int_{-3}^{1}(4y^{2}-28y + 49 - y^{4}-8y^{2}-16)dy\ &=\pi\int_{-3}^{1}(-y^{4}-4y^{2}-28y + 33)dy \end{align*} ]

Step4: Integrate term - by - term

[ \begin{align*} \int(-y^{4}-4y^{2}-28y + 33)dy&=-\frac{1}{5}y^{5}-\frac{4}{3}y^{3}-14y^{2}+33y+C \end{align*} ] [ \begin{align*} V&=\pi\left[-\frac{1}{5}y^{5}-\frac{4}{3}y^{3}-14y^{2}+33y\right]_{-3}^{1}\ &=\pi\left[\left(-\frac{1}{5}-\frac{4}{3}-14 + 33\right)-\left(\frac{243}{5}+36 - 126-99\right)\right]\ &=\pi\left[\left(-\frac{3 + 20}{15}-14 + 33\right)-\left(\frac{243}{5}+36-126 - 99\right)\right]\ &=\pi\left[\left(-\frac{23}{15}+19\right)-\left(\frac{243}{5}-189\right)\right]\ &=\pi\left[\frac{-23 + 285}{15}-\left(\frac{243-945}{5}\right)\right]\ &=\pi\left[\frac{262}{15}+\frac{702}{5}\right]\ &=\pi\left[\frac{262+2106}{15}\right]\ &=\frac{2368\pi}{15}\approx495.148 \end{align*} ]

Answer:

$495.148$