Line Integral Problem: Evaluating F(x,y) on Lower Half of Unit Circle

[Mr Noblet]

Homework Statement

Let F=x^{2}i+2xyj, and let C be the lower half of the unit circle, with perametrization r(t)=<cos(t),sin(t)>,\pi\leqt\leq\pi. Evaluate \ointF\cdotdr.

Homework Equations

The Attempt at a Solution

The first thing I tried to do was to find a function f(x,y) so that F=\nablaf

In order to do this I integrated the i portion of F with respect to x which gave me f(x,y)=\frac{1}{3}x^{3}+g(y).

Then to find out what g was I took the derivative with respect to y which just left me with g'(y). I could not think of how to get past this part. I tried doing it with the j first as well in which I first take the integral of the j part of F with respect to y and then the derivative with respect to x. This left me with f_{x}(x,y)=y^{2}+f'(x).

Basically I was trying to use the Fundamental Theorem of Line Integrals but I could not find a function f(x,y) so that F=\nablaf. Help would be appreciated.

 

[Avodyne]

It's not necessarily the case that F is the gradient of anything. If if happens to be, then the line integral is independent of the path, which makes it particularly easy; but if not, then you just have to do the integral.

 

[HallsofIvy]

Homework Statement

Let F=x^{2}i+2xyj, and let C be the lower half of the unit circle, with perametrization r(t)=<cos(t),sin(t)>,\pi\leqt\leq\pi.

Do you mean -\pi\le x\le 0? Or \pi\le x\le 2\pi which would work just as well.

Evaluate \ointF\cdotdr.

Homework Equations

The Attempt at a Solution

The first thing I tried to do was to find a function f(x,y) so that F=\nablaf

In order to do this I integrated the i portion of F with respect to x which gave me f(x,y)=\frac{1}{3}x^{3}+g(y).

Then to find out what g was I took the derivative with respect to y which just left me with g'(y). I could not think of how to get past this part. I tried doing it with the j first as well in which I first take the integral of the j part of F with respect to y and then the derivative with respect to x. This left me with f_{x}(x,y)=y^{2}+f'(x).

Basically I was trying to use the Fundamental Theorem of Line Integrals but I could not find a function f(x,y) so that F=\nablaf. Help would be appreciated.[/QUOTE]
You can't find such an f because x^2 dx+ 2xy dy is not an exact differential: (x^2)_y= 0 while (2xy)_x= 2y. Since those are not the same, there is no F such that grad F= x^2\vec{i}+ 2xy\vec{j}

With x= cos(t) and y= sin(t), dx= -sin(t)dt and dy= cos(t)dt x^2 dx+ 2xy dy= cos^2(t)(-sint dt)+ 2cos(t)sin(t)(cos(t) dt)= cos^2(t) sin(t) dt. It should be easy to integrate that directly.