I'm working through some old problem sets for a 4th class in calc (which usually deals with vector-calculus and PDEs), and encountered the following:
Find the volume generated by revolving the region bounded by the parabolas $y^2 = x$, $y^2 = 8x$, $x^2 = y$ and $x^2 = 8y$ about the $x$-axis by transforming cooridnates.
I can solve this by brute-force (separate the region to be rotated into $x$- and/or $y$-simple regions, find the volume contributed by each, then sum the volumes to find the total); however how can a change of coordinates be used to simplify the process?
Is there some way to make use of double integrals in application to volumes of revolution (otherwise I don't see what place such a question has in the standard curriculum for such a class, as these kinds of applications of integrals are introduced in the 2nd course...)
You can take the change of coordinates that make the limits of integration rectangular (see below).
Then, because of the symmetry of the region in the resulting space, you can use the distance of the centroid as the radius in the computation of the associated volume of revolution:
$u = \frac {x^2}{y}$ $v = \frac {y^2}{x}$
the region is now bounded by $u \in [1,8], v\in[1,8]$
$\frac {du}{dx} = \frac {2x}{y}, \frac {du}{dy} = \frac {-x^2}{2y^2}\\ \frac {dv}{dx} = \frac {-y^2}{2x^2}, \frac {dv}{dy} = \frac {2y}{x}$
Jacobian $= 4 - \frac 14 = \frac {15}{4}$
the area is $\frac {15}{4}\cdot 49$
The centroid $\frac {\iint u \ dA}{\iint dA}, \frac {\iint v \ dA}{\iint dA}$
$\frac 12 \frac {63\cdot 7}{49} = \frac {9}{2}, \frac {9}{2}$
$xy = uv \implies$ the centroid in $x,y$ coordinates is the same.
the volume of revolution is $2\pi r A = (2\pi)(\frac {9}{2})(\frac {15\cdot 49}{4})$