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I read about the famous Gaussian integral, i.e. $\int_0^\infty \exp(-x^2) \, dx$ and how to solve it. There are several approaches via polar coordinates or by defining some auxiliary functions that generate its square etc. However, how to compute the integral over a different, higher dimensional area?

Define the triangle $D := \{(x,y) \in \mathbb{R}^2 \mid 0\leq y \leq x \leq 1\}$. How would you calculate the Gaussian integral

$$ \int_D \exp(-x^2)\; d\lambda^2(x,y)$$

now?

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    This is easier than on a square or on the full space: integrate on $y$ first, then the resulting function of $x$ is an easy derivative. – 2017-01-30
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    You missed a method in the first paragraph. Easy substitution to turn it into the Gamma function. – 2017-01-30

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The particular case you mention is routine, but I think for some fairly simple geometric figures in the role of the region $D$ in $\int_D$, you'd have to use numerical methods. $$ \iint\limits_{x,y \,:\, 0 \le y\le x\le1} e^{-x^2} \, d(x,y) = \int_0^1 \left( \int_0^x e^{-x^2} \, dy \right) \, dx = \int_0^1 xe^{-x^2}\, dx = \int_0^1 e^{-x^2} \Big( x\,dx\Big). $$ Let $u=x^2$, so that $x\,dx = \dfrac 1 2 \, du$ and go on from there.