Answer to Question #315088 in Calculus for Nicz

Question

Find the volume of the solid obtained by rotating the region bounded by y = x^2 +1 and y = 9 – x^2 about y = -1.

Answer 1

Let’s go to new coordinates:

$\~x=x$;

$\~y=y+1$.

Now the volume of the solid can be obtained by rotating the region bounded by $\~y=\~x^2+2$ and $\~y=10-\~x^2$ about $\~y=0$.

Interval of integration:

$\~x^2+2= 10-\~x^2$

$\~x=\pm2$

$V=\int_{-2}^2 \pi((10-\~x^2)^2-(\~x^2+2)^2)d\~x=$

$\int_{-2}^2 \pi(96-24\~x^2)d\~x=\frac12 \int_{0}^2 \pi(96-24\~x^2)d\~x=$

$\frac12\pi (96\~x-8\~x^3)|_0^2=\pi(96-8\cdot4)=64\pi$

Answer: $V=64\pi$ .