How can i solve this double integral $$ \iint_D x(y+x^2)e^{y^2-x^4} dxdy $$ where $$D=\{(x,y) \in \mathbb{R}^2: x^2 \leq y \leq x^2+1, 2-x^2 \leq y \leq 3-x^2 \}? $$

  • Try changing the variables, using $u=x^2$ and $y$ as the new variables. – Harald Hanche-Olsen Feb 26 '12 at 20:18
  • I can write $D=\{(x,y)\in \mathbb{R}^2: 0 \leq y - x^2 \leq 1, 2 \leq y + x^2 \leq 3 \}$ and the integrand funcion as $f(x,y) =x(y+x^2)e^{(y-x^2)(y+x^2)}$. Sorry i wrote the wrong function first – Katy23 Feb 26 '12 at 20:32
  • With this notation i think that a possible change of variables is $u = y-x^2$ and $v=y+x^2$ but i don't know how to proceding – Katy23 Feb 26 '12 at 20:36
  • It looks to me as if because of the antisymmetry there is nothing to do, the integral is $0$. – André Nicolas Feb 26 '12 at 20:38
  • 2
    @Katy23: The region of integration is symmetric about the $y$-axis. The function you are integrating has (beside the $x^2$ stuff) an $x$ in front. So the function you are integrating is odd $(f(-x)=-f(x)$. The integral over the region to the left of the $y$ axis cancels the integral on the right. It is basically the same fact as $\int_{-a}^a x\cos(x^2)=0$. – André Nicolas Feb 26 '12 at 21:38

The region of integration is symmetric about the $y$-axis. The function you are integrating has (beside stuff that involves $x^2$), an $x$ in front.

So the function you are integrating is odd: $f(−x,y)=−f(x,y)$. The integral over the region to the left of the $y$-axis cancels the integral over the region to the right of the $y$-axis, so our answer is $0$. It is basically the same fact as $\int_{-a}^a x\cos(x^2)dx=0$.

That is enough, but if you wish, you can break up the region of integration into two halves along the $y$-axis, and for the integral over the region on the left, you can make the substitution $u=-x$. Don't integrate, just carry out the substitution process. After you are finished, replace the dummy variable $u$ of integration by $x$. Note that the region to the left of the $y$-axis has been carried by this process to the region on the right. All there is is a change of sign in the function you are integrating. Now add. We get $0$.

If the region of integration is, for example, the right half of your region, then we actually have to do some work! The substitution $u=x^2$ suggested by Harald Hanche-Olsen is then very helpful. You can also use more elaborate substitutions that involve both variables at once.

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