# Double integration with polar coordinates

• January 15th 2009, 12:50 PM
mitch_nufc
Double integration with polar coordinates
By changing to polar coordinates, evauluate the integral (where a>0):

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

Sorry I don't know latex, ill say it in words...

"...The integral from 0 to a from the square root of a square minus x squared of the function x^2+y^2 with respect to dydx"

I've got as far as changing my function to r^3 and changed my inner limits too a and 0, but im stuck on how to change my outer limits of a to 0 :S. I hope this makes sense
• January 15th 2009, 06:49 PM
mr fantastic
Quote:

Originally Posted by mitch_nufc
By changing to polar coordinates, evauluate the integral (where a>0):

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

Sorry I don't know latex, ill say it in words...

"...The integral from 0 to a from the square root of a square minus x squared of the function x^2+y^2 with respect to dydx"

I've got as far as changing my function to r^3 and changed my inner limits too a and 0, but im stuck on how to change my outer limits of a to 0 :S. I hope this makes sense

Draw the region of intergation in the xy-plane - it's the part of the circle of radius a and centre at origin that lies in the first quadrant.

In polar coordinates the region is clearly described by $0 \leq r \leq a$ and $0 \leq \theta \leq \frac{\pi}{2}$.
• January 16th 2009, 12:46 AM
mitch_nufc
But the way I solved changing the first set of limits is by setting y= sqrt(a^2 - x^2) thus y^2 + x^2 = a^2, then using the fact x=rcos and y=rsin i have r^2=a^2 (sin^2+cos^2=1) so r=a as a>0, in normal polars to find my angle i'd have tan = (y/x), so what would be the steps to find my angle in this case systematically? For other more complicated cases? Thanks
• January 16th 2009, 04:19 AM
mr fantastic
Quote:

Originally Posted by mitch_nufc
But the way I solved changing the first set of limits is by setting y= sqrt(a^2 - x^2) thus y^2 + x^2 = a^2, then using the fact x=rcos and y=rsin i have r^2=a^2 (sin^2+cos^2=1) so r=a as a>0, in normal polars to find my angle i'd have tan = (y/x), so what would be the steps to find my angle in this case systematically? For other more complicated cases? Thanks

I have told you what to do. Draw the region of integration. Use it to get the new integral terminals. That is what you do.
• January 16th 2009, 08:30 AM
HallsofIvy
Quote:

Originally Posted by mitch_nufc
By changing to polar coordinates, evauluate the integral (where a>0):

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

Sorry I don't know latex, ill say it in words...

"...The integral from 0 to a from the square root of a square minus x squared of the function x^2+y^2 with respect to dydx"

I've got as far as changing my function to r^3 and changed my inner limits too a and 0, but im stuck on how to change my outer limits of a to 0 :S. I hope this makes sense

What does the figure $0\le x\le a$, $0\le y\le \sqrt{a^2- x^2}$ look like?

Draw the boundaries x= 0, x= a, y= 0, $y= \sqrt{a^2- x^2}$ on a graph to see.