Line integral around an ellipse

[shoplifter]

Homework Statement

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.

Homework 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.

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?

 

[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.

 

[Altabeh]

Homework Statement

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.

Homework 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.

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?

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

AB

 

[Altabeh]

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.

A brilliant technical argument, but less mathematical.

 

[vela]

[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]

 

[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? :(

 

[vela]

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.

 

[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?

 

[vela]

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.)