Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.