Answer to Question #258929 in Calculus for Lie

Question

R is the region bounded by x = y

2 + 1 and x = 9 − y

2

and the axis

of revolution is x = 9.

Answer 1

$y^2+1=9-y^2$

$2y^2=8$

$y^2=4$

$y_1=-2, y_2=2$

$A(x)=\pi(outer\ radius)^2-\pi (inner\ radius)^2$

$=\pi (9-(y^2+1))^2-\pi(9-(9-y^2))^2$

$=\pi(81-18y^2-18+y^4+2y^2+1$

$-81+162-18y^2-81+18y^2-y^4)$

$=\pi(-16y^2+64)$

$V=\displaystyle\int_{-2}^2\pi(-16y^2+64)dy=32\pi\displaystyle\int_{0}^2\pi(-y^2+4)d$

$=32\pi\bigg[-\dfrac{y^3}{3}+4y\bigg]\begin{matrix} 2 \\ 0 \end{matrix}=32\pi(-\dfrac{8^3}{3}+8)$

$=\dfrac{512\pi}{3} (units^3)$