Evaluation of double integral – Polar Coordinates?

Question

I am trying to calculate the double integral $$\int_0^{\infty}\int_0^{\infty}e^{-2x^2-3y^2}\, dx\, dy$$ If both coefficients of $x^2$ and $y^2$ were equal then we could use polar coordinates.

Can we do that also in this case?

Then we would have $$\int\int e^{-r^2-r^2\cos^2\theta}\, r\, dr\, d\theta$$ But how can we continue from here? Or do we do something else here?

Answer 1

$$I=\int_0^\infty\int_0^\infty e^{-(2x^2+3y^2)}\,dx\,dy$$ now let $X=\sqrt{2}x\Rightarrow dx=\frac{dX}{\sqrt{2}}$ and $Y=\sqrt{3}y\Rightarrow dy=\frac{dY}{\sqrt{3}}$ so we get: $$I=\frac{1}{\sqrt{6}}\int_0^\infty\int_0^\infty e^{-(X^2+Y^2)}\,dX\,dY$$ now use your polar coordinates

Answer 2

If you already know the result from the Gaussian integral, $\int_0^\infty e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}$ (for $Re(a) > 0$) then you can simply break up the integral like this:

$$\int_0^\infty \int_0^\infty e^{-2x^2 - 3y^2} dx dy = \int_0^\infty e^{-2x^2} \left[ \int_0^\infty e^{-3y^2} dy \right] dx = \int^{\infty}_0 \sqrt{\frac\pi{3}}e^{-2x^2} dx = \frac\pi{\sqrt{6}}$$

which will match Henry's answer.