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 $y=-2$ and $y = 1$.

Step2: Use the method of cylindrical - shells

The radius of a shell is $r=7 - x$. For the region bounded by $x = 2 - y^{2}$ and $x = y$, using the formula for the volume of a solid of revolution by cylindrical - shells $V=2\pi\int_{a}^{b}r\cdot hdy$, where $a=-2$, $b = 1$, $r=7 - x$ (and since $x$ ranges from $x = y$ to $x = 2 - y^{2}$, the height $h=(2 - y^{2})-y$). We rewrite the volume formula as $V = 2\pi\int_{-2}^{1}(7 - x)[(2 - y^{2})-y]dy$. Since $x$ is a function of $y$, we have $V=2\pi\int_{-2}^{1}(7 - y)[(2 - y^{2})-y]dy$. Expand the integrand: [ \begin{align*} (7 - y)(2 - y^{2}-y)&=7(2 - y^{2}-y)-y(2 - y^{2}-y)\ &=14-7y^{2}-7y-(2y - y^{3}-y^{2})\ &=14-7y^{2}-7y - 2y + y^{3}+y^{2}\ &=y^{3}-6y^{2}-9y + 14 \end{align*} ]

Step3: Integrate

[ \begin{align*} V&=2\pi\int_{-2}^{1}(y^{3}-6y^{2}-9y + 14)dy\ &=2\pi\left[\frac{y^{4}}{4}-2y^{3}-\frac{9y^{2}}{2}+14y\right]_{-2}^{1}\ &=2\pi\left[\left(\frac{1^{4}}{4}-2\times1^{3}-\frac{9\times1^{2}}{2}+14\times1\right)-\left(\frac{(-2)^{4}}{4}-2\times(-2)^{3}-\frac{9\times(-2)^{2}}{2}+14\times(-2)\right)\right]\ &=2\pi\left[\left(\frac{1}{4}-2-\frac{9}{2}+14\right)-\left(4 + 16-18-28\right)\right]\ &=2\pi\left[\left(\frac{1 - 8 - 18+56}{4}\right)-\left(4 + 16-18-28\right)\right]\ &=2\pi\left[\frac{31}{4}-(-26)\right]\ &=2\pi\left(\frac{31 + 104}{4}\right)\ &=2\pi\times\frac{135}{4}\ &=\frac{135\pi}{2}\approx212.058 \end{align*} ]

Answer:

$212.058$