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)>,$$\pi$$$$\leq$$t$$\leq$$$$\pi$$. Evaluate $$\oint$$F$$\cdot$$dr.

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=$$\nabla$$f

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=$$\nabla$$f. 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)>,$$\pi$$$$\leq$$t$$\leq$$$$\pi$$.

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

Evaluate $$\oint$$F$$\cdot$$dr.

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=$$\nabla$$f

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=$$\nabla$$f. 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 [itex]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.