Calculate integral through a change of variables

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SUMMARY

The discussion focuses on calculating the integral of the function \( e^{-(x+y)^4}(x^2-y^2) \) over the region defined by \( D=\{ (x,y)\in\mathbb{R}^2:x+y< 1;0< y< x\} \) using a change of variables. The successful transformation involves setting \( u=x+y \) and \( v=x \), which simplifies the integral to \( 2\int_0^1\int_0^u e^{-u^4}uv \, dv \, du \). A critical error noted in the discussion was the miscalculation of the Jacobian, which should be \( \frac{1}{2} \) instead of \( 2 \).

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Homework Statement



Let
<br /> D=\{ (x,y)\in\mathbb{R}^2:x+y&lt; 1;0&lt; y&lt; x\}<br />
calculate \int_{D} e^{-(x+y)^4}(x^2-y^2)through an appropriate change of variables


Homework Equations



\int_{D} f *dxdy=\int_{D} f*Jacobian*dudv

The Attempt at a Solution



I've tried x+y=u and x-y=v which is a linear transformation so the jacobian is constant. However the integral becomes pretty ugly meanwhile and I can't solve it. What do i do?
 
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Try something more like

u=x+y
v=x
 
On the contrary, with that substitution, the integral becomes very nice! With u= x+ y. e^{-(x+y)^4}= e^{u^4} and with, also, v= x- y, x^2- y^2= (x- y)(x+ y)= uv so the integrand becomes e^{u^4}uv. Of course, the boundaries, x+y= 1, y= 0, and y= x, in the u, v system, become u= 1, v= 0, and u= v. The integral is
2\int_0^1\int_0^u e^{-u^4}uv dvdu

Now, do the first, very easy, integral with respect to v and then slap your forehead and cry "Of course"!
 
yeah right, did a mistake on the limits of integration of the v variable. I calculated the limiting points of the triangle in the uv plane and did a mistake in one point. Limits of v became v=u and v=1. This way was impossible.

Thanks but I think you did a mistake on the Jacobian. It is 1/2 not 2.
 

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