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 (or the washer method in terms of $y$).

The distance from a point $y$ to the axis of rotation $x = 7$ is $r=7 - x$. For the curves $x_1=y$ and $x_2=2 - y^{2}$, the outer - radius $R=7 - y$ and the inner - radius $r=7-(2 - y^{2})=5 + y^{2}$. The volume $V$ using the washer method with respect to $y$ is given by $V=\pi\int_{a}^{b}([R^{2}-r^{2}])dy$, where $a=-2$ and $b = 1$. $V=\pi\int_{-2}^{1}((7 - y)^{2}-(5 + y^{2})^{2})dy$. Expand the integrand: $(7 - y)^{2}=49-14y + y^{2}$ and $(5 + y^{2})^{2}=25 + 10y^{2}+y^{4}$. So $(7 - y)^{2}-(5 + y^{2})^{2}=49-14y + y^{2}-(25 + 10y^{2}+y^{4})=-y^{4}-9y^{2}-14y + 24$.

Step3: Integrate

$\int(-y^{4}-9y^{2}-14y + 24)dy=-\frac{1}{5}y^{5}-3y^{3}-7y^{2}+24y+C$.

Step4: Evaluate the definite integral

$V=\pi\left[-\frac{1}{5}y^{5}-3y^{3}-7y^{2}+24y\right]_{-2}^{1}$. $V=\pi\left[\left(-\frac{1}{5}(1)^{5}-3(1)^{3}-7(1)^{2}+24(1)\right)-\left(-\frac{1}{5}(-2)^{5}-3(-2)^{3}-7(-2)^{2}+24(-2)\right)\right]$. $V=\pi\left[\left(-\frac{1}{5}-3 - 7 + 24\right)-\left(\frac{32}{5}+24 - 28-48\right)\right]$. $V=\pi\left[\left(-\frac{1}{5}+14\right)-\left(\frac{32}{5}-52\right)\right]$. $V=\pi\left[\frac{-1 + 70}{5}-\frac{32-260}{5}\right]$. $V=\pi\left[\frac{69}{5}-\frac{-228}{5}\right]=\pi\left[\frac{69 + 228}{5}\right]=\frac{297\pi}{5}\approx186.646$.

Answer:

$186.646$