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 $y$ - axis)

The radius of a shell is $r=y + 3$ (distance from $x=-3$ to a point in the region). The height of the shell is $h=(y-(-2 + y^{2}))=y + 2 - y^{2}$. The volume formula for cylindrical - shells is $V=2\pi\int_{a}^{b}r\cdot hdy$. Here, $a=-1$, $b = 2$, $r=y + 3$, and $h=y + 2 - y^{2}$. So $V=2\pi\int_{-1}^{2}(y + 3)(y + 2 - y^{2})dy$.

Step3: Expand the integrand

$(y + 3)(y + 2 - y^{2})=y^{2}+2y-y^{3}+3y + 6-3y^{2}=-y^{3}-2y^{2}+5y + 6$.

Step4: Integrate term - by - term

$\int(-y^{3}-2y^{2}+5y + 6)dy=-\frac{1}{4}y^{4}-\frac{2}{3}y^{3}+\frac{5}{2}y^{2}+6y+C$.

Step5: Evaluate the definite integral

$V=2\pi\left[-\frac{1}{4}y^{4}-\frac{2}{3}y^{3}+\frac{5}{2}y^{2}+6y\right]_{-1}^{2}$. $V=2\pi\left[\left(-\frac{1}{4}(2)^{4}-\frac{2}{3}(2)^{3}+\frac{5}{2}(2)^{2}+6(2)\right)-\left(-\frac{1}{4}(-1)^{4}-\frac{2}{3}(-1)^{3}+\frac{5}{2}(-1)^{2}+6(-1)\right)\right]$. $V=2\pi\left[\left(-4-\frac{16}{3}+10 + 12\right)-\left(-\frac{1}{4}+\frac{2}{3}+\frac{5}{2}-6\right)\right]$. $V=2\pi\left[\left(18-\frac{16}{3}\right)-\left(-6+\frac{-3 + 8+30}{12}\right)\right]$. $V=2\pi\left[\frac{54 - 16}{3}-\left(-6+\frac{35}{12}\right)\right]$. $V=2\pi\left[\frac{38}{3}-\left(\frac{-72 + 35}{12}\right)\right]$. $V=2\pi\left[\frac{38}{3}+\frac{37}{12}\right]$. $V=2\pi\left[\frac{152+37}{12}\right]$. $V=2\pi\times\frac{189}{12}=\frac{189\pi}{6}=\frac{63\pi}{2}\approx98.960$.

Answer:

$98.960$