Understanding the Integral Limitations and Conversions in Polar Coordinates

[Freya]

Do the following integral: I= (double integral) dx dy x^2 with 0<x^2 + y^2<a^2 , y>0?
Could someone please give me some guidance as to what the inequality means? Should I convert to polar
coordinates so it gives me 0<r^2<a^2? Thank you.

 

[HallsofIvy]

0&lt; x^2+ y^2&lt; a^2, y> 0 is the half disk with center at (0, 0), radius a, above the x-axis. Yes, I think polar coordinates would be simplest although it would not be that hard to do it in Cartesian coordinates. Since y> 0, x^2+ y^2= a^2 is the same as y= \sqrt{a^2- x^2} so the integral would be \int_{x= -a}^a \int_{y= 0}^{\sqrt{a^2- x^2}} x^2 dy dx. What would it be in polar coordinates?

 

[Freya]

How does y >0 lead to x^2 + y^2 = a^2? Using polar coordinates the integral came to be (Sorry for lack of latex) integral of r^3 between 0 and a , times the integral of cos^2(theta) between 0 and pi. That came out to equal ((pi)a^4)/8. Does this sound right? Also could not seem to come to any answer using cartesian coordinates. thanks!

 

[Mark44]

How does y >0 lead to x^2 + y^2 = a^2?

It doesn't lead to $x^2 + y^2 = a^2$. It just limits the region of integration to the portion of the disk above the x-axis. .

Using polar coordinates the integral came to be (Sorry for lack of latex) integral of r^3 between 0 and a , times the integral of cos^2(theta) between 0 and pi. That came out to equal ((pi)a^4)/8. Does this sound right? Also could not seem to come to any answer using cartesian coordinates. thanks!