- #1

maxhersch

- 21

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## 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.