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Let $\Delta$ be a triangle with vertices $(0,0)$, $(0,1)$ and $(1,0)$ in ${\bf R}^2$. I want to compute $$I=\int\limits_\Delta x^2\mathrm{e}^{y^2}\;\mathrm{d}A.$$ This is what I've done so far: Note that the hypotenuse is given by the line $y=1-x$. Keep $x\in[0,1]$ fixed, then $y$ is between $1$ and $1-x$ this gives $I=\int_0^1\int_1^{1-x}x^2\mathrm{e}^{y^2}\;\mathrm{d}y\;\mathrm{d}x$, which can't be computed. But when $y\in[0,1]$ is fixed, $x$ is between $0$ and $1-y$, which gives $$I=\int\limits_0^1\int\limits_1^{1-y}x^2\mathrm{e}^{y^2}\;\mathrm{d}x\;\mathrm{d}y=\frac13\int\limits_0^1(1-y)^3\mathrm{e}^{y^2}\;\mathrm{d}y,$$ which gives the same trouble.

As you can see, I keep ending up with some sort of Gaussian integral, which is impossible to compute. Does anyone know how to compute this integral?

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    Quote: "then $y$ is in between $1$ and $1-x$" No, it's in between $0$ and $1-x$. You wrote the bounds incorrectly in a similar way on the 1st $\iint$ in the 2nd line of your question too. Here's a [picture](http://i.stack.imgur.com/nhRaG.gif) to keep track of the region of integration. W|A says the integral evaluates to something non-elementary, unfortunately, so I'm not sure if there's a nice answer here.2012-03-27
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    @anon. Oh, I see. Thank you for pointing it out. If I keep my $y$ fixed, is $x$ then between $1-y$ and $1$?, i.e. do I get the integral $\int_0^1 \int_{1-y}^1 ... dx dy?$2012-03-27
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    KLM, please think about the picture I linked. If $y$ is fixed it forms a horizontal line through the triangle. Does it look like where the shaded intersects the horizontal line, $x$ goes from the diagonal line $x=1-y$ to the vertical $x=1$ line?2012-03-27
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    @KLM: The integral is either $\int_0^1 dx \ \int_0^{1-x} dy$ or $\int_0^1 dy \ \int_0^{1-y} dx$. Notice that the hypotenuse is given by the formula $y=1-x$ (otherwise known as $x=1-y$). See anon's figure.2012-03-27

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