# Thread: Double Integral

1. ## Double Integral

Hey there, i am having trouble doing some questions on double integrals, ive done most of them.
Though im having trouble on question 2) ii) and question 3
i was wondering if someone could point me in the right direction on how to start evaluating these integrals.

thanks

2. Hello

For the second question, you can see this thread.

For the third one, the appropriate coordinates are the polar ones.

3. $\int e^{-x^2}~dx$ is called the Gaussian Integral. Even though the indefinite integral can't be evaluated, we can find the definite integral (from 0 to ∞) or (from -∞ to ∞).

Let $I = \int_0^{\infty}e^{-x^2}~dx$.

Then multiply it by itself: $I.I = \int_0^{\infty}e^{-x^2}~dx \int_0^{\infty}e^{-x^2}~dx$

Change x in the second integral to y:

Then multiply it by itself: $I.I = \int_0^{\infty}e^{-x^2}~dx \int_0^{\infty}e^{-y^2}~dy$

This is,

$I^2 = \int_0^{\infty}\int_0^{\infty}e^{-x^2-y^2}~dx~dy$

Change x-y to polar coordinates.

$x= r\cos\theta$
$y= r\sin\theta$
$dx~dy =r~dr~d\theta$

$I^2 = \int\int e^{-r^2}r~dr~d\theta$

$I^2 = \int d\theta \int e^{-r^2}r~dr$

The integration area is now $(0\leq\theta\leq \frac{\pi}{2})$ and $(0\leq r \leq \infty)$

$I^2 = \int_0^{\frac{\pi}{2}} d\theta \int_0^{\infty} e^{-r^2}r~dr$

Calculate the integral and take its square root to find I.