I have an intergral and im not sure how to find the limits of integration from cartesian to polar form 0<a<1
Integral from 1 to a ( integral from square root of (1-x^2) to 0) of F(x,y) dydx
Apparently the polar form integrals are suppose to be:
Integral from 1 to a (integral from cos-1(a/r) to 0) and im not sure where that came from...
Also if i have an integral from 1 to a(integral from root(1-x^2) to 0) of (x^2 -y^2)dydx i change it to polar form
integral 1 to a( integral from cos-1(a/r) of r(r^2cos^2theta - r^2sin^2thetha)dtheta dr
I attempted this integral by integrating with respect to r first then in respect to theta but am not sure if its right..any tips?
You find the limits of integration geometrically.
means that s ranges from 1 to a and, for each x, y ranges from [tex]\sqrt{1- x^2} to 0. leads to and, since y is the non-negative root, that is the upper half of the unit circle. Since for all x between -1 and 1 (and not defined otherwise), that integral is from the upper curve to the lower. That's why Danny said it would make more sense to reverse the limits of integration on the y integral. Of course, that multiplies the integral by -1: .
Now, x= 1 is the right end of that semi-circle while x= a, for a some number between -1 and 1. Points on the unit circle can be written (because then ) so if , , the arccosine. However, I disagree that the original integral is the same as what you give. If, for example, -1< a< 0, the left side of the region will be vertical line x= a, from y= 0 to . For from to , r will range from 0 to the distance to that vertical line, which is . For to r= 1.
So the integral should be .
Also if i have an integral from 1 to a(integral from root(1-x^2) to 0) of (x^2 -y^2)dydx i change it to polar form
integral 1 to a( integral from cos-1(a/r) of r(r^2cos^2theta - r^2sin^2thetha)dtheta dr
I attempted this integral by integrating with respect to r first then in respect to theta but am not sure if its right..any tips?