By changing to polar coordinates, evaluate the integral (where a>0):

int[a,0]int[sqrt(a^2 - x^2), 0](x^2 + y^2)dydx

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- Jan 1st 2009, 06:05 PMmitch_nufcdouble integration with polar coordinates
By changing to polar coordinates, evaluate the integral (where a>0):

int[a,0]int[sqrt(a^2 - x^2), 0](x^2 + y^2)dydx - Jan 1st 2009, 06:45 PMMush
May aswell post here!

Draw a picture of the region of the limits.

The lower limits are x = 0 and y = 0, which tells you that both x > 0 and y > 0, which means you're dealing with the 1st quadrant of the cartesian axis system.

Sketching the upper y limit

gives:

Circle with centre (0,0), radius a.

So from that sketch you should see that, in polar, the limits range from .

And the radius ranges from . Which converts to: