A really weird problem

The area of the circle at any point in the interval $x \in [0,20\pi]$ is $\pi[f(x)]^2=\pi(sinx+5)^2$ . Since $x$ varies from $0$ to $20\pi$ , the volume is given by $V=\displaystyle\int_{0}^{20\pi}\pi(sinx+5)^2{d}x=\pi\int_{0}^{20\pi}sin^2x{d}x+10\pi\int_{0}^{20\pi}sinxdx+25\pi\int_{0}^{20\pi}{d}x.$

$\pi\displaystyle\int_{0}^{20\pi}sin^2x{d}x=10\pi^2$

$10\pi\displaystyle\int_{0}^{20\pi}sinxdx=0$

$25\pi\displaystyle\int_{0}^{20\pi}{d}x=500\pi^2$

$\therefore V=10\pi^2+0+500\pi^2=510\pi^2$

We conclude that $a=510$ and $b=2$ . Therefore:

$a+b=512$