I was wondering, how do I find the volume of an area rotated about a diagonal axis (any $y=mx+c$) by using polar coordinate, where the function does not touch the axis of rotation at any point.
For example, how would I find the volume of the solid when $f(x)=\exp(x)$ is rotated about $y=x$, with the starting points $(0,1)$ and $(1,e)$?
I understand that I could convert $y=\exp(x)$ to $r\sin\theta = \exp(r\cos\theta)$, but what do I do about the starting and ending points?
Consider the vectors $u = s \pmatrix{-1\\1}$ and $v = s \pmatrix{1\\1}$, where $s = \frac{\sqrt{2}}{2}$ is chosen so that both $u$ and $v$ have unit length.
The point $(t, \exp(t))$ can be written as $$ \pmatrix{t\\ \exp(t)} = s(\exp(t) - t) u + s(\exp(t) + t) v, $$ as you can verify by direct computation. So in the $vu$ coordinate system, we can describe the exponential curve as $$ v(t) = s (\exp(t) + t) \\ u(t) = s (\exp(t) - t) $$ where $t$ ranges from $0$ to $1$.
Writing out an area formula for a volume of revolution of a parametric curve by doing the same sort of analysis done for volume of revolution of a function graph, we get $$ A = \int_{t_0}^{t_1} \pi [u(t)]^2 v'(t) ~dt $$ In your case, $t_0 = 0, t_1 = 1$, and $u$ and $v$ are given above, so you can take it from here, I'll bet.
NB: As a quick sanity check, for $vu$-coordinates, we can take a function graph like $u = f(c)$ and make it a parametric curve by choosing $v(t) = t$ and $u(t) = f(v(t)) = f(t)$, and when we plug into the formula above, the $v'$ term becomes $1$, so it reduces the the area formula for a function graph.