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 \}? $
How to solve this double integral with a not-normal domain
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0Perfect! You are right – 2012-02-26
1 Answers
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.