Question involving fundamental theorem of line integrals

[Thunderdan700]

Homework Statement

a) Use the fundamental theorem of line integrals to evaluate the line integral:
∫(2x/(x^2+y^2)^2)dx+(2y/(x^2+y^2)^2)dy (over C) Where C is the arc of the circle (x-4)^2+(y-5)^2=25 taken clockwise from (7,9) to (0,2). Explain why the fundamental theorem can be applied.
b) Obtain the same result by parameterizing the curve and evaluating the resulting integral

Homework Equations

fundamental theorem of line integrals :
∫(del)f * dr = f(r(b))-f(r(a))

The Attempt at a Solution

I'm supposed to find the function for which the function inside the integral is the gradient. I have:df/dx=(2x/(x^2+y^2)^2) =>> f=-(x^2+y^2)^-1 and df/dy=(2y/(x^2+y^2)^2) =>> f=-(x^2+y^2)^-1, so f(x,y)=-(x^2+y^2)^-1. As far as parameterizing the circle, I have x=5cost+4 and y=5sint+5. I'm completely stuck at this point. I don't see how to get values of t that yield those points on the circle, and I'm confused further because the circle is traversed in a clockwise direction. Any help would be appreciated, thanks.

 

[LCKurtz]

Draw a graph of your circle, with another set of axes through its center. Your parameter t is the polar coordinate angle for this second set of axes. For (7,9), look at the little xy triangle from the center (4,5) of the circle to (7,9). You can read sin(t), cos(t), and tan(t) for that angle off that little 3-4-5 triangle, and you can express t in terms of whichever of the inverse trig functions is appropriate. Similarly for the other point. Use a negative angle taking care because it isn't in the range of the inverse trig functions. Once you get those angles do your line integral from the first value of t to the second for clockwise. Or go the other way and use a minus sign in front of the integral.