Rotate an area around a diagonal line. I know how to find the volume of the figure formed when you rotate a $2$-dimensional area around a horizontal or vertical line, but what if it were a diagonal line instead?
For example:
Rotate the area between the curve $y=x^2$ and $y=x$ around the line $y=x$.
That should create a diagonal "football" shape, starting at the origin.
Anyways, the way I thought to go about the problem was to rotate the equation $45°$ to the right, and then solve the problem as if it were rotated around the x-axis. So I decided to convert $y=x^2$ to a polar equation, add $\frac{pi}{4}$ to $\theta$, and then turn it back into a rectangular equation. That worked, and it gave me this equation, which gives a diagonal parabola opening toward the first quadrant:
$\frac{\sqrt{2}}{2}(y+x) = \frac{1}{2}xy(x^2+y^2)$
Only problem is that I need to isolate $y$, such that I can solve this problem like an ordinary rotation problem, and that seems impossible. What should I do to solve the problem? I don't know much linear algebra, but could that be used to solve this instead?


Convert $y=x^2$ to a polar equation

$$r\sin\theta = r^2\cos^2 \theta$$
$$r\sin\theta = r^2\cos^2 \theta \rightarrow \hbox{ (add }\frac{\pi}{4}\hbox{ to }\theta\hbox{)}$$
$$r\sin (\theta + \frac{\pi}{4}) = r^2 \cos^2(\theta + \frac{\pi}{4})$$
$$\frac{\sqrt{2}}{2}r\left(\sin \theta + \cos \theta\right) = \frac{1}{2} \left(r\cos\theta -r\sin \theta\right)^2 \rightarrow\hbox{ (convert back) }$$
$$\frac{\sqrt{2}}{2}(y+x) = \frac{1}{2} (x-y)^2$$ 
which is not $\frac{\sqrt{2}}{2}(x+y)=\frac{1}{2}xy(x^2+y^2)$ anyway.
Thus, it must be a mistake somewhere :)
$\rho((x^2,x),y-x=0)$ can be done much simpler: $\rho=|x^2-x|\cdot \frac{\sqrt{2}}{2}$, so you just have to $\pi\int \rho^2(t) dt$ where $t=\frac{\sqrt{2}}{2}(x+y)=\frac{\sqrt{2}}{2}(x^2+x)$.
Edit:$\frac{\sqrt{2}}{2}(x+y)=\frac{1}{2}xy(x^2+y^2)$ itself leads to an irreducible cubic.
