# Thread: Change of variables in 3 dimensions

1. ## Change of variables in 3 dimensions

(First post, hoorah!)

Hi there - I'm revising for my exams in a few weeks and decided to look at some of the older past papers for my uni course, but they tend to deviate off the syllabus a fair bit so I haven't been taught how to approach a problem like this systematically, I'm using mostly guesswork and logic...

(i) 'The domain S in the (x,y) plane is bounded by $y=x$, $y=ax (0 \leq a \leq 1)$ and $xy^2=1 (x,y \geq 0)$. Find a transformation
$u=f(x,y) v=g(x,y)$
such that S is transformed into a rectangle in the (u,v) plane.'

I figured I may as well aim to transform into the unit square if I could, since it's only a short step translating from a rectangle in the (u,v) plane to the unit square. The only problem which struck me here was the fact a rectangle needs 4 'bounds' whereas the given area in the (x,y) plane has only 3 bounds. Through, as I said, mostly fairly unconfident guesswork, I came up with $u=\frac{xy^2(y-ax)}{\frac{1}{\sqrt{x}}-ax}, v=\frac{xy^2(y-x)}{\frac{1}{\sqrt{x}}-x}$ - would this work? It gives u=1 and v=1 along the $xy^2=1$ boundary and 0 along the y=x and y=ax boundaries, so would that be okay? In general, what's a good method for approaching this sort of question systematically?

(ii) 'Evaluate $\int_D \frac{y^2z^2}{x} \, dx \, dy \, dz$, where D is the region bounded by

$y=x$, $y=zx$, $xy^2=1$ $(x,y \geq 0)$

and the planes $z=0$, $z=1$.'

Obviously a change of variables is the way they want you to go, and I'm happy with the actual process of changing variables, the Jacobian, the volume integral etc, but how do I actually work out -what- variables to change to? It's not like it's a symmetrical area which points towards spherical polars or anything like that, so I'm not sure how to proceed!

Thanks very much for the help!

2. For (i), I would write the inequalities as $a\leqslant\frac yx\leqslant1$ and $0\leqslant xy^2\leqslant1$. So if $u(x,y) = \frac xy$ and $v(x,y) = xy^2$ then S corresponds to the rectangle $a\leqslant u\leqslant1,\ 0\leqslant v\leqslant1$ in the (u,v)-plane.

Then $uv = y^3$ and $v/u^2 = x^3$, from which $y^2/x = (uv)^{2/3}(v/u^2)^{-1/3} = u^{4/3}v^{1/3}$. So for (ii) you just have to integrate $u^{4/3}v^{1/3}z^2$ (multiplied by a suitable Jacobian determinant) over the region $z\leqslant u\leqslant1,\ 0\leqslant v\leqslant1,\ 0\leqslant z\leqslant1$.