# $\iiint e^{-x^2-2y^2-3z^2}dV$

I was given this question in class but I don't understand how to do it.

Evaluate the triple integral in $\mathbb{R}^3$ of $\iiint e^{-x^2-2y^2-3z^2}dV$.

The hint was to use the idea that $\int e^{-x^2}dx = \sqrt \pi$, which I understand, but I don't get how to apply it here...

If anyone can help I would appreciate it.

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This instantly separates into a product of three integrals, each of a function of one variable. –  Michael Hardy Mar 24 '13 at 4:51

Hint: By Fubini's theorem, we can rewrite the integral as $$\iiint e^{-(x^2+2y^2+3z^2)}\,dV=\left(\int e^{-x^2}\,dx\right)\left(\int e^{-2y^2}\,dy\right)\left(\int e^{-3z^2}\,dz\right).$$Now you know the value for the first one, and try a particular substitution for the second and third integrals (each requires a separate substitution) to get the integral into a familiar form.

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Got it! Thanks a lot. –  user68203 Mar 24 '13 at 16:00