question find the volume of the solid obtained by rotating the region bounded by x = -6 + y² and x = -y…

Explanation:

Step1: Find intersection points

Set $-6 + y^{2}=-y$. Rearrange to $y^{2}+y - 6=0$. Factor: $(y + 3)(y - 2)=0$. So $y=-3$ and $y = 2$.

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

The outer radius $R(y)$ and inner radius $r(y)$ for rotation about $x=-8$. The distance from the axis $x = - 8$ to a point $x$ is $d=x + 8$. So $R(y)=(-6 + y^{2})+8=y^{2}+2$ and $r(y)=-y + 8$.

Step3: Apply the volume formula

The volume formula for the washer method when rotating about a vertical line is $V=\pi\int_{a}^{b}(R^{2}(y)-r^{2}(y))dy$. Here $a=-3$, $b = 2$, $R(y)=y^{2}+2$, $r(y)=-y + 8$. [ \begin{align*} V&=\pi\int_{-3}^{2}((y^{2}+2)^{2}-(-y + 8)^{2})dy\ &=\pi\int_{-3}^{2}(y^{4}+4y^{2}+4-(y^{2}-16y + 64))dy\ &=\pi\int_{-3}^{2}(y^{4}+4y^{2}+4 - y^{2}+16y - 64)dy\ &=\pi\int_{-3}^{2}(y^{4}+3y^{2}+16y - 60)dy \end{align*} ]

Step4: Integrate term - by - term

$\int y^{4}dy=\frac{1}{5}y^{5}$, $\int3y^{2}dy=y^{3}$, $\int16ydy = 8y^{2}$, $\int-60dy=-60y$. [ \begin{align*} V&=\pi\left[\frac{1}{5}y^{5}+y^{3}+8y^{2}-60y\right]_{-3}^{2}\ &=\pi\left[\left(\frac{1}{5}(2)^{5}+(2)^{3}+8(2)^{2}-60(2)\right)-\left(\frac{1}{5}(-3)^{5}+(-3)^{3}+8(-3)^{2}-60(-3)\right)\right]\ &=\pi\left[\left(\frac{32}{5}+8 + 32-120\right)-\left(-\frac{243}{5}-27 + 72 + 180\right)\right]\ &=\pi\left[\left(\frac{32}{5}+40-120\right)-\left(-\frac{243}{5}+225\right)\right]\ &=\pi\left[\left(\frac{32}{5}-80\right)-\left(-\frac{243}{5}+225\right)\right]\ &=\pi\left[\frac{32-400}{5}-\frac{-243 + 1125}{5}\right]\ &=\pi\left[\frac{-368}{5}-\frac{882}{5}\right]\ &=\pi\left[\frac{-368 - 882}{5}\right]\ &=\pi\left[\frac{-1250}{5}\right]\ &=250\pi\approx 785.398 \end{align*} ]

Answer:

$785.398$