let $r$ be the region bounded by the curves $x = y^{2}-2$ and $x = 2 - y^{2}$. what is the volume of the…

Explanation:

Step1: Find intersection points

Set $y^{2}-2 = 2 - y^{2}$. Then $2y^{2}=4$, so $y^{2}=2$ and $y =-\sqrt{2},\sqrt{2}$.

Step2: Use the method of cylindrical - shells (or the distance - formula for volume of revolution about a vertical line)

The distance from a point $x$ to the line $x = 5$ is $d=5 - x$. The outer - radius $R=5-(y^{2}-2)=7 - y^{2}$ and the inner - radius $r = 5-(2 - y^{2})=3 + y^{2}$. The volume $V=\pi\int_{-\sqrt{2}}^{\sqrt{2}}((7 - y^{2})^{2}-(3 + y^{2})^{2})dy$. Expand the integrand: $(7 - y^{2})^{2}-(3 + y^{2})^{2}=(49-14y^{2}+y^{4})-(9 + 6y^{2}+y^{4})=40-20y^{2}$.

Step3: Integrate

$V=\pi\int_{-\sqrt{2}}^{\sqrt{2}}(40-20y^{2})dy$. Since $\int_{-\sqrt{2}}^{\sqrt{2}}(40-20y^{2})dy=\int_{-\sqrt{2}}^{\sqrt{2}}40dy-\int_{-\sqrt{2}}^{\sqrt{2}}20y^{2}dy$. $\int_{-\sqrt{2}}^{\sqrt{2}}40dy=40y\big|{-\sqrt{2}}^{\sqrt{2}}=40(\sqrt{2}-(-\sqrt{2})) = 80\sqrt{2}$. $\int{-\sqrt{2}}^{\sqrt{2}}20y^{2}dy=20\times\frac{y^{3}}{3}\big|_{-\sqrt{2}}^{\sqrt{2}}=\frac{20}{3}(2\sqrt{2})=\frac{40\sqrt{2}}{3}$. $V=\pi\left(80\sqrt{2}-\frac{40\sqrt{2}}{3}\right)=\frac{200\sqrt{2}\pi}{3}$.

Answer:

$\frac{200\sqrt{2}\pi}{3}$