Volume of rotated region (integration) Let $T$ be a right-angled triangular region with vertices $(0,−b)$,$(1,0)$ and $(0,a)$ where $a$ and $b$ are positive numbers. When $T$ is rotated about the line $x=2$, it generates a solid with volume $V=\dfrac{410\pi}{27}$
Find $a$ and $b$.
Really having trouble with this one. Is $r=(2-x)$ and $h=(a+b)(1-x)$, and how do I integrate that?
Then, how do I display the result as a result of '$a$' and '$b$'?

Using Pappus's Theorem, $V=A(2\pi\rho)=\frac{1}{2}(a+b)(2\pi(2-\frac{1}{3}))=\pi(a+b)\frac{5}{3}=\frac{410}{27}\pi$, 
$\hspace{.3 in}$ so $a+b=\frac{82}{9}$.  
By the Pythagorean Theorem, 
$a^2+1+b^2+1=(a+b)^2,\;\;\;$ so $2=2ab\implies ab=1\implies b=\frac{1}{a}$.
Then $a+\frac{1}{a}=\frac{82}{9}\implies \frac{a^2+1}{a}=\frac{82}{9}\implies9a^2-82a+9=0\implies(9a-1)(a-9)=0$,
so $a=\frac{1}{9}$ and $b=9,\;\;\;$ or $a=9$ and $b=\frac{1}{9}$.
