# Evaluating a double integral

1. Jan 22, 2012

### Lucy Yeats

1. The problem statement, all variables and given/known data

Evaluate the integral (x^2+y^2)arctan(y/x) for 0<y<a and 0<x<(a^2-y^2)^0.5.

2. Relevant equations

3. The attempt at a solution

I tried changing the order of integration to get the integral (x^2+y^2)arctan(y/x) for 0<x<a and 0<y<(a^2-x^2)^0.5. I noticed that this was a quarter of a circle.

I tried then taking x^2 out of the dy integral and into the dx one. The dy integral is then (1+(y/x)^2)arctan(y/x). I'm not sure what to do now.

Last edited: Jan 23, 2012
2. Jan 22, 2012

### Lucy Yeats

Do you use the fact that x^2+y^2=a^2? I can then take the constant a^2 in front of the integral, so I only need to integrate arctan(y/x).dy.

3. Jan 23, 2012

### Staff: Mentor

Is the last inequality supposed to be 0 < x < √(a2 - y2)?

Rather than just changing the order of integration, I think the best plan is change to a polar integral.

4. Jan 23, 2012

### Lucy Yeats

Yes, I've changed that now- thanks for pointing out the error.

So is it a^2∫θ.dθ∫r.dr?

How would I change the limits?

5. Jan 23, 2012

### Lucy Yeats

So would the limits be 0<r<a and -π/2<θ<π/2?

6. Jan 23, 2012

### Lucy Yeats

Would someone mind checking whether these limits are correct? :-)

7. Jan 23, 2012

### I like Serena

Hi Lucy Yeats!

It should be x^2+y^2=r^2.
It is not equal to a^2, since you integrate (x,y) over the surface of the circular disk and not just the boundary.

Your angle θ should run from 0 to pi/2, since you only integrate the first quadrant.
For θ<0 you would get negative y, but your problem statement says y>0.

8. Jan 27, 2012

### Lucy Yeats

I've got it now, thanks everyone! :-)