Change of Variables for Double Integral

1. Aug 10, 2011

cjc0117

Update: I figured out how to solve the problem. Nevermind.

1. The problem statement, all variables and given/known data

Use a suitable change of variables to evaluate the double integral:

$\int^{1}_{0}\int^{3-x}_{2x}(y-2x)e^{(x+y)^{3}}dydx$

2. Relevant equations

$\frac{\partial(x,y)}{\partial(u,v)}=( \frac{\partial x}{\partial u} )(\frac{\partial y}{\partial v})-(\frac{\partial x}{\partial v})(\frac{\partial y}{\partial u})$

$dA=|\frac{\partial(x,y)}{\partial(u,v)}|dudv$

3. The attempt at a solution

I made the following change of variables:

$u=x+y$

$v=2x$

(At first I tried u=y-2x and v=x+y but that didn't work)

Next, I solved for x and y:

$x=\frac{v}{2}$

$y=\frac{2u-v}{2}$

I found that the Jacobian is:

$\frac{\partial(x,y)}{\partial(u,v)}=-\frac{1}{2}$

And therefore...

$dA=\frac{1}{2}dudv$

Next I changed the boundaries:

$x=0\rightarrow v=0$

$x=1\rightarrow v=2$

$y=3-x\rightarrow u=3$

$y=2x\rightarrow u=\frac{3}{2}v$

Therefore, the new region S on the uv-plane is the triangle described by:

$0\leq v\leq 2$

$\frac{3}{2}v\leq u\leq 3$

The original integrand becomes:

$(y-2x)e^{(x+y)^{3}}\rightarrow \frac{2u-3v}{2}e^{u^{3}}$

So the new integral is:

$\frac{1}{4}\int^{2}_{0}\int^{3}_{\frac{3}{2}v}(2u-3v)e^{u^{3}}dudv$

I used wolfram alpha to find out that the integral is equal to $\frac{1}{18}(e^{27}-1)$, which is the correct answer, but I am not sure how to evaluate the integral. Do I have to use another change of variables? Is there a better change of variables I could have used in the beginning?

Thank you. Any help is much appreciated.

Update:

Okay, I figured out how to solve the integral. I'll post the rest of the solution in case anybody else who has this same question ever happens across this post. What I did was I switched the order of integration so that the integral becomes:

$\frac{1}{4}\int^{3}_{0}\int^{\frac{2}{3}u}_{0}(2u-3v)e^{u^{3}}dvdu$

$=\frac{1}{4}\int^{3}_{0}(2uv-\frac{3}{2}v^{2})e^{u^{3}}]^{\frac{2}{3}u}_{0}du$

$=\frac{1}{6}\int^{3}_{0}u^{2}e^{u^{3}}du$

Use substitution with $w=u^{3}$ and $dw=3u^{2}du\rightarrow \frac{1}{3}dw=u^{2}du$:

$\frac{1}{18}\int^{27}_{0}e^{w}dw$

$=\frac{1}{18}e^{w}]^{27}_{0}$

$=\frac{1}{18}(e^{27}-1)$

Last edited: Aug 10, 2011