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 '$b$'?
 A: Your rotated volume is made of two parts ($y>0$ and $y<0$): both are made of a cylinder minus a truncated cone. Thus the volume is easy to compute, if you know $a$ and $b$: this gives you an equation between $a$ and $b$.
For the second equation, you know your triangle is right angled: the right angle must be at vertex $(1,0)$, so $(a+b)^2=1+a^2+1+b^2$ and $ab=1$.
Then solve for $a$ and $b$.

More precisely, the "upper" volume is, with $R=2$ and $r=1$ (see here),
$$V_{upper}=\pi R^2a-\frac a3(\pi R^2+\pi r^2+\pi R r)=\frac{a\pi}3(3R^2-R^2-r^2-R r)$$
Likewise, the lower volume is:
$$V_{lower}=\pi R^2b-\frac b3(\pi R^2+\pi r^2+\pi R r)=\frac{b\pi}3(3R^2-R^2-r^2-R r)$$
So the total volume is
$$V=\frac{(a+b)\pi}3(2R^2-r^2-R r)=\frac{5(a+b)\pi}3=\frac{410\pi}{27}$$
So
$$a+b=\frac{82}{9}=9+\frac{1}{9}$$
$$ab=1$$
Hence $a=9$ and $b=\frac{1}{9}$, or the other way around.

If you insist on computing an integral, you can write
$$V=\int_{-b}^a S(h)\mathrm{d}h$$
Where $S(h)$ is the area of the intersection of your volume and the plane $y=h$, hence, for $h>0$
$$S(h)=4\pi-\pi\left(1+\frac ha\right)^2$$
The first term is the area of a disc of radius $2$. The second is the area of a disc of radius $1+\frac ha$ (use for example Thales' theorem to see why).
For $h<0$, you have $S(h)=4\pi-\pi\left(1-\frac hb\right)^2$
Hence
$$V=4a\pi-\pi\int_0^a \left(1+\frac ha\right)^2\mathrm{d}h+4b\pi-\pi\int_{-b}^0 \left(1-\frac hb\right)^2\mathrm{d}h$$
$$V=4a\pi-\pi\int_0^a \left(1+\frac ha\right)^2\mathrm{d}h+4b\pi-\pi\int_0^b \left(1+\frac hb\right)^2\mathrm{d}h$$
$$V=4a\pi-\pi\left(a+\frac{a^2}{a}+\frac{a^3}{3a^2}\right)+4b\pi-\pi\left(b+\frac{b^2}{b}+\frac{b^3}{3b^2}\right)=\frac{5(a+b)\pi}{3}$$
Of course you find the same volume as in the formula above.
A: 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 $b=\frac{1}{9}$.
