Jacobian Change of Variables Question

[themadhatter1]

Homework Statement

Evaulate the integral making an appropriate change of variables.

\int\int_R(x+y)e^{x^2-y^2}dA where R is the parallelogram enclosed by the lines x-2y=0, x-2y=4, 3x-y=1, 3x-y=8 .

Homework Equations

The Attempt at a Solution

I'm not sure what change of variables I should make. The way the region R is defined suggests that I should make the substitution u=x-2y, v=3x-y. Which maps the region r into a square s which is a simple region to integrate over. However, solving for x and y you obtain x=(1/5)(2v-u), y= (1/5)(v-3u). Changing the integral using these substitutions yields

\frac{1}{5}(3v-4u)e^{\frac{1}{25}(-8u^2+2uv+3v^2)} which is not integrable.

Likewise, if you select a substitution which makes the integrand simple, say u=x+y and v=x-y you obtain a parallelogram as the region s which is not simple to integrate over. Am I missing something?

 

[Dick]

Making the integrand simple is usually more important than making the borders simple. I would go with the parallelogram.

 

[spamiam]

How can you rewrite the exponent?

 

[themadhatter1]

How can you rewrite the exponent?

you can rewrite the exponents as (x-y)(x+y). Using the substitution u=x-y and y=x+y you have a parallelogram with bounds v=3u,v=3u-8,v=-2u+1,v=-2u+8. Therefore, the area can be represented by the following integrals

\int_{0.2}^{1.6}\int_{-2u+1}^{3u}ve^{uv}dvdu+\int_{1.6}^{1.8}\int_{-2u+1}^{-2u+8}ve^{uv}dvdu+\int_{1.8}^{3.2}\int_{3u-8}^{-2u+8}ve^{uv}dvdu

These integrals are very difficult to work with and you can't find an anti-derivative for the outside integral. I suppose at least I can approximate the value with a calculator now, but there has to be an easier way which yields an exact solution that you don't have to integrate over a parallelogram.

 

[Dick]

I didn't work the whole problem out. I usually do, but I don't like this one any better than you do. But if you integrate dv first then the antiderivative of v*exp(uv)=(uv-1)*exp(uv)/u^2. That doesn't look good. But if you integrate du first the antiderivative is just exp(uv). That looks much better. I'm not saying that will make it easy, but it certainly should make it better. Try changing the order of integration.