# Line Integral

1. Jul 9, 2012

### portuguese

Consider that the curve C (IR3) is the intersection between x2+2z2=2 and y=1.

Calculate the line integral:

Note: The curve C is traversed one time in the counter-clockwise direction (seen from the origin of IR3).

My attempt:

http://i.imgur.com/5EsJO.png

Can someone check? Thanks! :)

Last edited by a moderator: Jul 9, 2012
2. Jul 10, 2012

### haruspex

If I have the question right, a lot of simplification is possible.
Given y = 1, the integral wrt y is zero, and y can be set to 1 elsewhere.
I also notice there's integral x.dz - integral z.dx. Hmmm... what do you think that might simplify to? Think of the geometric meaning, or of integration by parts.

3. Jul 10, 2012

### portuguese

But it is still correct?

4. Jul 10, 2012

### haruspex

The answer looks right. I haven't verified all the working (because it can all be done in couple of lines).

5. Jul 10, 2012

### portuguese

Thanks! Btw, I understand your simplifications, but I have to do this way... ;)

6. Jul 10, 2012

### HallsofIvy

Staff Emeritus
You have curve C defined in an xyz- coordinate system but you have u, y, and z in your integral. Did you intend
$$\int_C yzdx+ x^2dy- (x+ z)dz$$?

Assuming that, since y= 1, a constant, dy= 0 so we can ignore the middle term.

I would be inclined to parameterize the curve by $x= \sqrt{2}cos(\theta)$, $y= 1$, $z= sin(\theta)$.

Then $dx= -\sqrt{2}sin(\theta)d\theta$ and $dz= cos(\theta)d\theta$.

The integral becomes
$$\int_0^{2\pi} sin(\theta)(-\sqrt{2}sin(\theta)d\theta)- (\sqrt{2}cos(\theta)+ sin(\theta))cos(\theta)d\theta$$
$$= \int_0^{2\pi}[-\sqrt{2}(sin^2(\theta)+ cos^2(\theta))- \sqrt{2}sin(\theta)cos(\theta)]d\theta$$
$$= -\sqrt{2}\int_0^{2\pi}[1+ sin(\theta)cos(\theta)]d\theta$$

The answer you give, t/2- (1/4)sin(2t), can't possibly be correct because it is a function of t and there was not even a "t" in the problem! Did you intend to evaluate that at t= 0 and $2\pi$?

Last edited: Jul 10, 2012
7. Jul 10, 2012

### haruspex

The way I read it, the equations involving t are explanatory notes. The answer it gives earlier is "= 0".

8. Jul 10, 2012

### HallsofIvy

Staff Emeritus
Thank you. I had to keep scrolling from side to side!

9. Jul 10, 2012

### portuguese

10. Jul 10, 2012

### HallsofIvy

Staff Emeritus
I've gone back and edited my first response, replacing the incorrect "2"s with $\sqrt{2}$. However, I still do not get 0 as the integral.

Last edited: Jul 10, 2012
11. Jul 10, 2012

### portuguese

I found my mistake. I wrote:
It's not 0, you're right. Thanks a lot! ;)

12. Jul 10, 2012

### haruspex

And I've found mine .
As HofI, I reasoned that we can get rid of the y terms, leaving:
$\int_C zdx- (x+ z)dz = \int_C (zdx-xdz) + zdz$
Around a closed loop, zdz goes to zero (at each z, a dz cancels a -dz).
$\int_C zdx$ and $\int_C xdz$ are each the area inside the loop. So I thought these cancelled too. But I overlooked that one will effectively run the opposite direction around the loop, so will add, not cancel. So the whole reduces to twice the area of the ellipse.