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

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 $r=(x+4)$ and the height of the shell is $h = (-y)-(-2 + y^{2})=-y^{2}-y + 2$. The volume formula for rotating about a vertical line using cylindrical - shells is $V=\int_{a}^{b}2\pi r h\ dy$. Here, $a=-2$, $b = 1$, $r=(y^{2}-2 + 4)=y^{2}+2$ (for the parabola) and $r=-y + 4$ (for the line), and $h=-y^{2}-y + 2$. [ \begin{align*} V&=2\pi\int_{-2}^{1}[(y^{2}+2)-(-y + 4)][-y^{2}-y + 2]\ dy\ &=2\pi\int_{-2}^{1}(y^{2}+y - 2)(-y^{2}-y + 2)\ dy\ &=-2\pi\int_{-2}^{1}(y^{2}+y - 2)^{2}\ dy\ &=-2\pi\int_{-2}^{1}(y^{4}+2y^{3}-3y^{2}-4y + 4)\ dy \end{align*} ]

Step3: Integrate term - by - term

[ \begin{align*} \int_{-2}^{1}(y^{4}+2y^{3}-3y^{2}-4y + 4)\ dy&=\left[\frac{y^{5}}{5}+\frac{2y^{4}}{4}-y^{3}-2y^{2}+4y\right]_{-2}^{1}\ &=\left(\frac{1}{5}+\frac{1}{2}-1 - 2+4\right)-\left(-\frac{32}{5}+8 + 8-8-8\right)\ &=\left(\frac{2 + 5}{10}-1 - 2+4\right)-\left(-\frac{32}{5}\right)\ &=\left(\frac{7}{10}+1\right)+\frac{32}{5}\ &=\frac{7 + 10}{10}+\frac{64}{10}\ &=\frac{81}{10} \end{align*} ]

Step4: Calculate the volume

$V=-2\pi\times\frac{81}{10}=-\frac{81\pi}{5}\approx - 50.894$ (but volume is non - negative), so $V\approx50.894$.

Answer:

$50.894$