Volume of area bounded by $x^2+y^2=a^2$ and $x+y=a$, revolving $y=x+2a$ Find the volume of area bounded under by $x^2+y^2=a^2$ and above by $x+y=a$ revolving $y=x+2a$, where $a>0$ and sketch it.
Here is what I tried.
For a sketch, I have tried, but not ensure if that true. Anyone can send me the correct sketch?
From my sketch I have, I can get the volume by integrating the difference between the radius of outer circle and the radius inner circle. Let $r_s$ and $r_b$ be the radius of inner circle and the radius of outer circle respectively. I can get them by find distance from $(-\frac{a}{2},\frac{3}{2})$ to $(0,a)$ for $r_s$ and distance from $(-\frac{a}{2},\frac{3a}{2})$ to $(a,0)$ for $r_b$, since the center of circle is on
$(-\frac{a}{2},\frac{3a}{2})$.
So, I have 
$V = \int(0,a-\frac{a}{\sqrt(2)}) ((x+2a-r_b)^2-(x+2a-r_s)^2)$
and then.
Is it correct? I get stuck and I'm not ensure my way is correct. Thanks for help in advanced.


The volume is equivalent to that of the area bounded between $x^2+y^2=a^2$ and $x=\frac a{\sqrt2}$ revolving around  $x=\sqrt2a$. Then, with disk integral method, the volume is 
\begin{align}
V &= \int_{-\frac a{\sqrt2}}^{\frac a{\sqrt2}}\pi(r_1^2-r_2^2)dy\\
&=\pi  \int_{-\frac a{\sqrt2}}^{\frac a{\sqrt2}}
\left[\left(\sqrt2 a - \frac a{\sqrt2}\right)^2 - \left(\sqrt2 a - \sqrt{a^2-y^2}\right)^2\right]dy\\
&=\frac{\pi}{3\sqrt2}(3\pi-8)a^3
\end{align}
