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 terms of $y$)

The radius of a shell is $r=(x+4)$. For the curves $x_1=-2 + y^{2}$ and $x_2=-y$, the height of the shell $h=(x_1 - x_2)=(-2 + y^{2}+y)$. The volume formula using the shell method about the line $x=-4$ is $V = 2\pi\int_{a}^{b}r\cdot h\ dy$, where $a=-2$, $b = 1$, and $r=(y + 4)$. So $V=2\pi\int_{-2}^{1}(y + 4)(y^{2}+y - 2)\ dy$.

Step3: Expand the integrand

$(y + 4)(y^{2}+y - 2)=y^{3}+y^{2}-2y+4y^{2}+4y - 8=y^{3}+5y^{2}+2y - 8$.

Step4: Integrate term - by - term

$\int(y^{3}+5y^{2}+2y - 8)dy=\frac{y^{4}}{4}+\frac{5y^{3}}{3}+y^{2}-8y+C$.

Step5: Evaluate the definite integral

$V = 2\pi\left[\frac{y^{4}}{4}+\frac{5y^{3}}{3}+y^{2}-8y\right]_{-2}^{1}$ $=2\pi\left[\left(\frac{1}{4}+\frac{5}{3}+1 - 8\right)-\left(\frac{16}{4}-\frac{40}{3}+4 + 16\right)\right]$ $=2\pi\left[\left(\frac{3 + 20+12 - 96}{12}\right)-\left(4-\frac{40}{3}+4 + 16\right)\right]$ $=2\pi\left[\frac{-61}{12}-\left(24-\frac{40}{3}\right)\right]$ $=2\pi\left[\frac{-61}{12}-\frac{72 - 40}{3}\right]$ $=2\pi\left[\frac{-61}{12}-\frac{32}{3}\right]$ $=2\pi\left[\frac{-61-128}{12}\right]$ $=2\pi\left(\frac{-189}{12}\right)=\frac{-189\pi}{6}=-\frac{63\pi}{2}\approx - 98.960$ (the negative sign just indicates the orientation of the calculation, we take the absolute value). $V\approx98.960$

Answer:

$98.960$