1. Jun 25, 2009

### Knissp

EDIT:
SORRY, I didn't read the directions. It says, "Answer the following short questions: If true, justify, if false give a counterex-
ample."

I'm certain that this question is one of the "false" ones, which is why I was so confused. LOL

1. The problem statement, all variables and given/known data
Let f(x, y, z) = y - x. Then the line integral of grad(f) around the unit circle in the xy plane is $$\pi$$, the area of the circle.

2. Relevant equations
A line integral of a vector field which is the gradient of a scalar field is path independent.

3. The attempt at a solution
I had two ways of solving:

Method 1
Fundamental theorem of line integrals:
$$\oint_C \nabla f dr = 0$$ around a closed curve C.
Path independence guarantees that the line integral of grad(f) depends only on the initial and final points, which are the same on the unit circle.

Method 2
$$grad(f) = <-1, 1, 0>$$

Parametrize the unit circle C by:
$$x(t) = cos(t) 0 \leq t \leq 2\pi$$
$$y(t) = sin(t) 0 \leq t \leq 2\pi$$
$$z=0$$
$$x'(t) = -sin(t)$$
$$y'(t) = cos(t)$$

$$\int_C grad(f) dr$$
$$= \int <-1, 1, 0><dx, dy, dz>$$
$$= \int -dx + \int dy$$
$$= \int_0^{2\pi} -\frac{dx}{dt}dt +\int_0^{2\pi} \frac{dy}{dt}dt$$
$$= \int_0^{2\pi} sin(t) dt + \int_0^{2\pi} cos(t) dt$$

$$= 0$$

Using either method, I get 0 as the answer. The question suggests that the answer should be $$\pi$$. Am I missing something?

2. Jun 26, 2009