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 $3 - y^{2}=-2y$, which is $y^{2}-2y - 3 = 0$. 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 $r=9 - x$. For the curves $x_1 = 3 - y^{2}$ and $x_2=-2y$, the volume formula $V=\pi\int_{a}^{b}[(r_1)^{2}-(r_2)^{2}]dy$, where $r_1 = 9-(3 - y^{2})=6 + y^{2}$ and $r_2=9 + 2y$.

Step3: Set up the integral

$V=\pi\int_{-1}^{3}[(6 + y^{2})^{2}-(9 + 2y)^{2}]dy$. Expand the integrand: $(6 + y^{2})^{2}=36+12y^{2}+y^{4}$ and $(9 + 2y)^{2}=81 + 36y+4y^{2}$. The integrand is $y^{4}+12y^{2}+36-(4y^{2}+36y + 81)=y^{4}+8y^{2}-36y - 45$.

Step4: Evaluate the integral

$\int(y^{4}+8y^{2}-36y - 45)dy=\frac{y^{5}}{5}+\frac{8y^{3}}{3}-18y^{2}-45y+C$. $V=\pi\left[\frac{y^{5}}{5}+\frac{8y^{3}}{3}-18y^{2}-45y\right]_{-1}^{3}$. First, substitute $y = 3$: $\frac{3^{5}}{5}+\frac{8\times3^{3}}{3}-18\times3^{2}-45\times3=\frac{243}{5}+72 - 162-135=\frac{243}{5}+72-(162 + 135)=\frac{243}{5}+72 - 297=\frac{243}{5}-225=\frac{243-1125}{5}=-\frac{882}{5}$. Then substitute $y=-1$: $\frac{(-1)^{5}}{5}+\frac{8\times(-1)^{3}}{3}-18\times(-1)^{2}-45\times(-1)=-\frac{1}{5}-\frac{8}{3}-18 + 45=-\frac{3 + 40}{15}+27=-\frac{43}{15}+27=\frac{-43 + 405}{15}=\frac{362}{15}$. $V=\pi\left(-\frac{882}{5}-\frac{362}{15}\right)=\pi\left(\frac{-2646 - 362}{15}\right)=\pi\times\frac{-3008}{15}\approx - 630.539$. But volume is non - negative, so $V\approx630.539$.

Answer:

$630.539$