Double Integral of a Quarter Circle: Evaluating (x^2+y^2)arctan(y/x)

[Lucy Yeats]

Homework Statement

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

Homework Equations

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.

 

[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.

 

[Mark44]

Homework Statement

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

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

Homework Equations

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.

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

 

[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?

 

[Lucy Yeats]

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

 

[Lucy Yeats]

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

Thanks in advance.

 

[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.

 

[Lucy Yeats]

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