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

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 $y=-3$ and $y = 1$.

Step2: Use the method of cylindrical - shells

The radius of a shell is $r=7 - x$. For the curves $x = 3 - y^{2}$ and $x = 2y$, the volume $V$ using the formula $V=2\pi\int_{a}^{b}r\cdot hdy$, where $a=-3$, $b = 1$, and $h=(3 - y^{2}-2y)$. The radius $r = 7 - x$, and since we are integrating with respect to $y$, $r=7 - 2y$ (we can also use the other $x$ - expression in the radius formula, the result will be the same). So $V = 2\pi\int_{-3}^{1}(7 - 2y)[(3 - y^{2})-2y]dy$. Expand the integrand: [ \begin{align*} (7 - 2y)(3 - y^{2}-2y)&=7(3 - y^{2}-2y)-2y(3 - y^{2}-2y)\ &=21-7y^{2}-14y-(6y - 2y^{3}-4y^{2})\ &=21-7y^{2}-14y - 6y + 2y^{3}+4y^{2}\ &=2y^{3}-3y^{2}-20y + 21 \end{align*} ]

Step3: Integrate

[ \begin{align*} V&=2\pi\int_{-3}^{1}(2y^{3}-3y^{2}-20y + 21)dy\ &=2\pi\left[\frac{2y^{4}}{4}-\frac{3y^{3}}{3}-\frac{20y^{2}}{2}+21y\right]{-3}^{1}\ &=2\pi\left[\frac{1}{2}y^{4}-y^{3}-10y^{2}+21y\right]{-3}^{1} \end{align*} ] Evaluate at the upper and lower limits: [ \begin{align*} &\text{When }y = 1:\frac{1}{2}(1)^{4}-(1)^{3}-10(1)^{2}+21(1)=\frac{1}{2}-1 - 10 + 21=\frac{1 - 2-20 + 42}{2}=\frac{21}{2}\ &\text{When }y=-3:\frac{1}{2}(-3)^{4}-(-3)^{3}-10(-3)^{2}+21(-3)=\frac{81}{2}+27-90 - 63=\frac{81 + 54-180 - 126}{2}=-\frac{171}{2} \end{align*} ] [ \begin{align*} V&=2\pi\left[\frac{21}{2}-\left(-\frac{171}{2}\right)\right]\ &=2\pi\times\frac{21 + 171}{2}\ &=192\pi\approx603.186 \end{align*} ]

Answer:

$603.186$