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

Explanation:

Step1: Find intersection points

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

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

The radius of a shell is $r=-y+8$ (distance from $x=-y$ to $x = - 8$) and $r=-2 + y^{2}+8=y^{2}+6$ (distance from $x=-2 + y^{2}$ to $x=-8$). The height of the shell $h=( - 2 + y^{2})-(-y)=y^{2}+y - 2$. The volume formula for the method of cylindrical - shells about a vertical line $x = a$ is $V = 2\pi\int_{c}^{d}(x - a)h\mathrm{d}y$. Here, $V=2\pi\int_{-2}^{1}((-8)-x)h\mathrm{d}y=2\pi\int_{-2}^{1}((-8)-(-y))((-2 + y^{2})-(-y))\mathrm{d}y+2\pi\int_{-2}^{1}((-8)-(-2 + y^{2}))((-2 + y^{2})-(-y))\mathrm{d}y$. First, expand $(y + 8)(y^{2}+y - 2)=y^{3}+y^{2}-2y+8y^{2}+8y - 16=y^{3}+9y^{2}+6y - 16$. Second, expand $(-y^{2}-6)(y^{2}+y - 2)=-y^{4}-y^{3}+2y^{2}-6y^{2}-6y + 12=-y^{4}-y^{3}-4y^{2}-6y + 12$. Then $V = 2\pi\int_{-2}^{1}(y^{3}+9y^{2}+6y - 16 - y^{4}-y^{3}-4y^{2}-6y + 12)\mathrm{d}y=2\pi\int_{-2}^{1}(-y^{4}+5y^{2}-4)\mathrm{d}y$.

Step3: Integrate term - by - term

$\int(-y^{4}+5y^{2}-4)\mathrm{d}y=-\frac{1}{5}y^{5}+\frac{5}{3}y^{3}-4y+C$. Evaluate the definite integral: [ \begin{align*} &2\pi\left[-\frac{1}{5}y^{5}+\frac{5}{3}y^{3}-4y\right]_{-2}^{1}\ =&2\pi\left[\left(-\frac{1}{5}(1)^{5}+\frac{5}{3}(1)^{3}-4(1)\right)-\left(-\frac{1}{5}(-2)^{5}+\frac{5}{3}(-2)^{3}-4(-2)\right)\right]\ =&2\pi\left[\left(-\frac{1}{5}+\frac{5}{3}-4\right)-\left(\frac{32}{5}-\frac{40}{3}+8\right)\right]\ =&2\pi\left[\left(-\frac{3 + 25-60}{15}\right)-\left(\frac{96 - 200 + 120}{15}\right)\right]\ =&2\pi\left[\frac{-38}{15}-\frac{16}{15}\right]\ =&2\pi\times\frac{-54}{15}\ =&\frac{-108\pi}{15}\ =&\frac{-36\pi}{5}\approx - 22.619 \end{align*} ] Since volume is non - negative, $V\approx22.619$.

Answer:

$22.619$