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 $2 - y^{2}=y$, so $y^{2}+y - 2=0$. Factoring gives $(y + 2)(y - 1)=0$. Thus $y=-2$ and $y = 1$.

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

The outer radius $R(y)=4 - y$ and the inner radius $r(y)=4-(2 - y^{2})=2 + y^{2}$. The volume formula for rotating about a vertical line using the washer method with respect to $y$ is $V=\pi\int_{a}^{b}([R(y)]^{2}-[r(y)]^{2})dy$. Here, $a=-2$, $b = 1$, $[R(y)]^{2}=(4 - y)^{2}=16-8y + y^{2}$ and $[r(y)]^{2}=(2 + y^{2})^{2}=4 + 4y^{2}+y^{4}$. Then $[R(y)]^{2}-[r(y)]^{2}=16-8y + y^{2}-(4 + 4y^{2}+y^{4})=-y^{4}-3y^{2}-8y + 12$.

Step3: Calculate the integral

$V=\pi\int_{-2}^{1}(-y^{4}-3y^{2}-8y + 12)dy$. Integrating term - by - term: $\int -y^{4}dy=-\frac{1}{5}y^{5}+C$, $\int-3y^{2}dy=-y^{3}+C$, $\int - 8y dy=-4y^{2}+C$, $\int12dy = 12y+C$. Evaluating the definite integral: [ \begin{align*} V&=\pi\left[-\frac{1}{5}y^{5}-y^{3}-4y^{2}+12y\right]_{-2}^{1}\ &=\pi\left[\left(-\frac{1}{5}(1)^{5}-(1)^{3}-4(1)^{2}+12(1)\right)-\left(-\frac{1}{5}(-2)^{5}-(-2)^{3}-4(-2)^{2}+12(-2)\right)\right]\ &=\pi\left[\left(-\frac{1}{5}-1 - 4+12\right)-\left(\frac{32}{5}+8 - 16-24\right)\right]\ &=\pi\left[\left(\frac{-1 - 5-20 + 60}{5}\right)-\left(\frac{32 + 40-80 - 120}{5}\right)\right]\ &=\pi\left[\frac{34}{5}-\left(-\frac{128}{5}\right)\right]\ &=\pi\times\frac{34 + 128}{5}\ &=\frac{162\pi}{5}\ &\approx101.788 \end{align*} ]

Answer:

$101.788$