# Jacobian Change of Variables Question

## 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 .

## 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?

## Answers and Replies

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

How can you rewrite the exponent?

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 cant 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