# Double Integration in Polar Coordinates

1. Dec 19, 2016

### maxhersch

1. The problem statement, all variables and given/known data
Integrate by changing to polar coordinates:

$\int_{0}^6 \int_{0}^\sqrt{36-x^2} tan^{-1} \left( \frac y x \right) \, dy \, dx$

2. Relevant equations
$x = r \cos \left( \theta \right)$
$y = r \sin \left( \theta \right)$

3. The attempt at a solution

So this is a quarter of a circle in the first quadrant of an xy-coordinate system with a radius of 6. So the bounds for the integral in polar coordinates will be r from 0 to 6 and $\theta$ from 0 to $\frac{ \pi }{2}$ .

$tan^{-1} \left( \frac y x \right)$ is just $\theta$ so the integral becomes:

$\int_{0}^6 \int_{0}^ \frac{ \pi }{2} \theta \, d \theta \, dr$

I worked this out and got $\frac{3 \pi ^2}{4}$ . I checked it on Wolfram Alpha and got the same result. I plugged the original integral into Wolfram Alpha and got $\frac{9 \pi ^2}{4}$ so somehow I lost a factor of 3 somewhere. Clearly I integrated properly so I must have made a mistake converting to polar coordinates.

Any help would be appreciated. Thanks.

2. Dec 19, 2016

### rock.freak667

In your conversion to polar the elemental area in Cartesian is dA = dy*dx

However in polar it should be r*dr*dθ

3. Dec 19, 2016

### eys_physics

You have forgot a factor of $r$. The area element $dxdy$ should be replaced by $rdrd\theta$, which is obtained from the Jacobian for the transformation to polar coordinates. I hope this helps.

4. Dec 19, 2016

### maxhersch

Yep, just remembered that right as your replied. Thanks

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