# Line integral around an ellipse

1. Jan 9, 2010

### shoplifter

1. The problem statement, all variables and given/known data
What is $$\int_{\gamma} xy dx + x^2 dy$$ in each of the following cases?

1. $$\gamma$$ is the lower half of the curve $$2x^2 + 3y^2 = 8$$, traveled from $$(2,0)$$ to $$(-2,0)$$.

2. $$\gamma$$ is the full curve $$2x^2 + 3y^2 = 8$$, traveled counterclockwise.

2. Relevant equations

The line integral formula, I suppose. The fact that the integral can be expressed as the dot product of the vector field $$(xy, x^2)$$ with the unit tangent vector to the curve can also be helpful.

3. The attempt at a solution

I parametrized the curves for (1) and (2) in different ways, viz.

1. $$x = t, y = -2\sqrt{\frac{2}{3}\left(1 - \frac{t^2}{4}\right)}$$.

2. $$x = 2\cos{\theta}, y = 2\sqrt{2/3}\sin{\theta}$$.

Then standard integration rules, but I came up with 0 for both the answers. Am I correct?

2. Jan 9, 2010

### boboYO

For no.1, the xy dx part is equal to 0, by symmetry: for every point (x,y) [which contributes xy dx] on the curve, there is another point (-x,y) [which contributes -xy dx] on the opposite side of the y axis. dx is always positive on this curve, so they cancel each other out.

We can also see that the x^2 dy part is equal to 0: because for the first half of the curve, dy is negative, for the second half, dy is positive, while x^2 is positive and mirrored. so the first half cancels out the second half again.

So 0 seems right.

2nd question follows easily from the first by splitting it up into 2 integrals: first around the bottom half and then around the top half of the ellipse. We already know from the first question that the bottom half = 0, and by symmetry the top half must be 0 too. 0+0=0.

Symmetry arguments are good for checking your work.

3. Jan 9, 2010

### Altabeh

I checked your answers and they are correct. Here is a graphical analysis of the problem in both cases:
http://img30.imageshack.us/img30/8343/intego.jpg [Broken]

AB

Last edited by a moderator: May 4, 2017
4. Jan 9, 2010

### Altabeh

A brilliant technical argument, but less mathematical.

5. Jan 9, 2010

### vela

Staff Emeritus
[STRIKE]Note that $$d\Phi(x,y)=xydx+x^2dy$$ is an exact differential, so the integrals depend only on the endpoints. (I'll leave it to you to find $$\Phi(x,y)$$.) The second integral is then trivially zero because the start and end points are the same. The first integral is equal to $$\Phi(-2,0)-\Phi(2,0)=0$$.[/STRIKE]

Last edited: Jan 9, 2010
6. Jan 9, 2010

### shoplifter

many thanks, guys! i can't find $$\Phi(x)$$ though, i keep just missing it by a scalar factor. it would seem that $$\Phi = x^2y$$ would work, but it just misses it. poincare's lemma doesn't seem to help either. am i missing something obvious? :(

7. Jan 9, 2010

### vela

Staff Emeritus
Why does $$\Phi(x,y)=x^2y$$ not work? You could add on a constant, obviously, but that wouldn't make a difference in evaluating the integral.

8. Jan 9, 2010

### shoplifter

oh, i thought because $$d(x^2y) = 2xy dx + x^2 dy$$. how do we get rid of the scalar multiple of 2? it isn't an additive constant, right?

9. Jan 9, 2010

### vela

Staff Emeritus
Oh, I'm sorry. I misled you. For some reason I thought there was a two there in the original problem. You're right. It's not exact, so you have to do the integral by hand. (And they were both zero when I did them earlier.)