How to calculate limits of integration?

[kilroymcb]

When converting from rectangular to polar coordinates in a double integral over a region?

Say, you're given the function:
Double integral over R of (x+y) dA where R is the region to the left of the y axis between the circles (x^2)+(y^2) =1 and (x^2)+(y^2) = 4.

I know that when I convert the function to polar coordinates, I'll have double integral of the region (rcos(theta) + rsin(theta)) d(theta)d(r). Also, I know for one of the limits of integration it would be easier to make the inequality 1<=(x^2)+(y^2)<=4 and therefore 1<=r^2<=4 and that means one of the limits of integration is 1<= r <= 2. How would I find the other? I guess I missed a major point in class.

Thanks

 

[tkhunny]

There is no substitute for simply thinking about it.

Left of the y-axis. Try \(\displaystyle [\frac{\pi}{2},\frac{3\pi}{2}]\)

Feel free to exploit symmetries: Twice the value obtained from \(\displaystyle [\frac{\pi}{2},\pi]\)

 

[Deleted member 4993]

When converting from rectangular to polar coordinates in a double integral over a region?

Say, you're given the function:
Double integral over R of (x+y) dA where R is the region to the left of the y axis between the circles (x^2)+(y^2) =1 and (x^2)+(y^2) = 4.

I know that when I convert the function to polar coordinates, I'll have double integral of the region

(rcos(theta) + rsin(theta)) r d(theta)d(r). <--- Don't forget that 'r' in there.

Also, I know for one of the limits of integration it would be easier to make the inequality 1<=(x^2)+(y^2)<=4 and therefore 1<=r^2<=4 and that means one of the limits of integration is 1<= r <= 2. How would I find the other? I guess I missed a major point in class.

Thanks