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

Explanation:

Step1: Find intersection points

Set $- 2 + y^{2}=y$, then $y^{2}-y - 2=0$. Factoring gives $(y - 2)(y+1)=0$. So $y=-1$ and $y = 2$.

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

The radius of a shell is $r=y + 5$ (distance from $x=-5$ to a vertical line at a given $y$ - value). The height of the shell is $h=y-(-2 + y^{2})=y + 2 - y^{2}$. The volume formula for the method of cylindrical - shells is $V=2\pi\int_{a}^{b}r\cdot hdy$. Here, $a=-1$, $b = 2$, $r=y + 5$, and $h=y + 2 - y^{2}$. So $V=2\pi\int_{-1}^{2}(y + 5)(y + 2 - y^{2})dy$.

Step3: Expand the integrand

$(y + 5)(y + 2 - y^{2})=y^{2}+2y-y^{3}+5y + 10-5y^{2}=-y^{3}-4y^{2}+7y + 10$.

Step4: Integrate term - by - term

$\int(-y^{3}-4y^{2}+7y + 10)dy=-\frac{1}{4}y^{4}-\frac{4}{3}y^{3}+\frac{7}{2}y^{2}+10y+C$.

Step5: Evaluate the definite integral

$V=2\pi\left[-\frac{1}{4}y^{4}-\frac{4}{3}y^{3}+\frac{7}{2}y^{2}+10y\right]_{-1}^{2}$ $=2\pi\left[\left(-\frac{1}{4}(2)^{4}-\frac{4}{3}(2)^{3}+\frac{7}{2}(2)^{2}+10(2)\right)-\left(-\frac{1}{4}(-1)^{4}-\frac{4}{3}(-1)^{3}+\frac{7}{2}(-1)^{2}+10(-1)\right)\right]$ $=2\pi\left[\left(-4-\frac{32}{3}+14 + 20\right)-\left(-\frac{1}{4}+\frac{4}{3}+\frac{7}{2}-10\right)\right]$ $=2\pi\left[\left(30-\frac{32}{3}\right)-\left(\frac{-3 + 16+42 - 120}{12}\right)\right]$ $=2\pi\left[\frac{90 - 32}{3}-\frac{-65}{12}\right]$ $=2\pi\left[\frac{58}{3}+\frac{65}{12}\right]$ $=2\pi\left[\frac{232 + 65}{12}\right]$ $=2\pi\times\frac{297}{12}=\frac{297\pi}{6}\approx155.509$

Answer:

$155.509$