find the volume of the solid obtained by rotating the region bounded by x = 3 - y² and x = -2y about the…

Explanation:

Step1: Find intersection points

Set $3 - y^{2}=-2y$, then $y^{2}-2y - 3 = 0$. Factoring gives $(y - 3)(y+1)=0$. So $y=-1$ and $y = 3$.

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

The radius of a shell is $r=5 - x$. For the curves $x_1 = 3 - y^{2}$ and $x_2=-2y$, the volume formula using the shell method $V=\int_{a}^{b}2\pi r h\ dy$, where $a=-1$, $b = 3$, $r = 5 - x$, and $h=(3 - y^{2})-(-2y)$. We rewrite the volume formula as $V = 2\pi\int_{-1}^{3}[(5-(3 - y^{2}))- (5 + 2y)]dy=2\pi\int_{-1}^{3}[(2 + y^{2})-(5 + 2y)]dy=2\pi\int_{-1}^{3}(y^{2}-2y - 3)dy$.

Step3: Integrate

$\int(y^{2}-2y - 3)dy=\frac{1}{3}y^{3}-y^{2}-3y+C$. Evaluating the definite - integral: [ \begin{align*} 2\pi\left[\frac{1}{3}y^{3}-y^{2}-3y\right]_{-1}^{3}&=2\pi\left[\left(\frac{1}{3}(3)^{3}-(3)^{2}-3(3)\right)-\left(\frac{1}{3}(-1)^{3}-(-1)^{2}-3(-1)\right)\right]\ &=2\pi\left[(9 - 9-9)-\left(-\frac{1}{3}-1 + 3\right)\right]\ &=2\pi\left[-9-\left(\frac{-1 - 3+9}{3}\right)\right]\ &=2\pi\left[-9-\frac{5}{3}\right]\ &=2\pi\left[\frac{-27 - 5}{3}\right]\ &=2\pi\times\left(-\frac{32}{3}\right)\ &=\frac{64\pi}{3}\approx67.021 \end{align*} ]

Answer:

$67.021$