Double Integration in Polar Coordinates

[maxhersch]

Homework Statement

Integrate by changing to polar coordinates:

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

Homework Equations

$x = r \cos \left( \theta \right)$
$y = r \sin \left( \theta \right)$

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.

 

[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

 

[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.

 

[maxhersch]

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

However in polar it should be r*dr*dθ

Yep, just remembered that right as your replied. Thanks