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Calculus/Arctan integration

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Expert: - 2/9/2010


Question
Dear Scotto,
I am trying to figure out how to do a double integral that needs to be converted to polar coordinates.  The book has the question written as ∫∫arctan(y/x)dA where R= {(x,y) 1≤ x^2 +y^2≤4, 0≤y≤x}.  I set the first intergral as y=1∫y=4 and the second as x=0∫x=pi.  I don't know if I set the integrals correctly but I can't figure out how to start integrating arctan(y/x).  I tired replacing y with rsinθ and x with rcosθ and I don't know how to integrate the y over x with arctan.  Any suggestions are greatly appreciated.  
Thanks,
Britta

Answer
What I do know is that y/x = tan(θ), so arctan(y/x) is θ.

The value of θ varies from 0 to 2π.

The value of R² varies from 1 to 4, and in this case R is positive, so R goes from 1 to 2.
If we allowed R to be negative as well, would only vary from 0 to π,
for the area from π to 2π would have been accounted for already.

When converting ∫∫ f(x,y) dx dy to polar, it becomes ∫∫ f(r,θ) r dr dθ.

This then becomes the ∫∫ θ² dr dθ, where r goes from 1 to 2 and θ goes from 0 to 2π.

Integrate with respect to r and just add an r.
Integrate with respect to θ and just make it θ³/3.
After this, put in the limits of r and θ into rθ³/3.

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