Is it possible to integrate with respect to a defined segment instead of a given axis? Today I was solving a calculus exercise related to solids of revolution and I came up with, I think, a very difficult question. The exercise, as shown below in the link, asks for the volume of a solid when you rotate some areas around a couple of given segments.

The exercise becomes pretty simple if you rotate any of the given areas with respect to $\overline{OA}$ or $\overline{OC}$ , x and y axis respectively, but what has actually made me thinking was if you tried to rotate, for example, $\mathcal{R}_3$ around $\overline{OB}$. That would be the solid created by rotating $y=\sqrt[4]{x}$ with respect the segment $\overline{OB}$. Now I have the conception that in order to integrate this, the best way is to rotate the function to get an integral like $\int_{a}^{b} f(x)\, dx$ and be able to do it as usual with respect to x or y axis. 
Alright, so the question I have is: Can you integrate with respect to a custom segment instead of the traditional x and y integration? No matter how hard it is, I just want to find out if it is possible.
 A: It can be done indeed and you're correct about rotating the function and then taking the integral like you normally would but I'd do it so that instead of rotating the function, I would rotate the axis. In this case you would take the integral with respect to the "new $x$-axis". Let's see how it would be done with your particular example.
Let's begin by naming our new axis: Let $z$ and $w$ represent the "new $x$ and $y$ axis" respectively. Since the $z$-axis must go trough the line segment $\overline{OB}$, it can be represented with the equation $y=x$. We can choose the $w$-axis to be represented by $y=-x$. We say:
$$z=y-x\\w=y+x$$
EDIT: 
This transformation of the axis also scales them by $\sqrt{2}$. To counteract that, we must multiply by $1/\sqrt{2}$:
$$z=(y-x)/\sqrt{2}\\w=(y+x)/\sqrt{2}.$$
Now we have to express the equation $y=\sqrt[4]{x}\Leftrightarrow y^4 = x$ in terms of $z$ and $w$.
We do it by first solving a system of equation for $x$ and $y$.
$$
\begin{cases}
z=(y-x)/\sqrt{2}
\\
w=(y+x)/\sqrt{2}
\end{cases}
\Leftrightarrow x=\frac{w-z}{\sqrt{2}} \quad \textrm{and} \quad y=\frac{w+z}{\sqrt{2}}
$$
Now we just need to substitute.
$$\begin{align}
\label{eqn:eqlabel}
\begin{split}
y^4&=x
\\
(\frac{w+z}{\sqrt{2}})^4&=\frac{w-z}{\sqrt{2}}
\end{split}
\end{align}$$
Now you would in theory solve for $w$ and integrate $\pi\int_0^{\sqrt{2}}w^2dz$ but the equation for $w$ is so huge that it's practically impossible to integrate by hand (See Wolfram|Alpha)...
You can check that all of this is correct by plotting $(\frac{w+z}{\sqrt{2}})^4=\frac{w-z}{\sqrt{2}}$. You will see that the resulting curve is basically $y=\sqrt[4]{x}$ but rotated by $45^{\circ}$(Plot here)
