Calculating Line Integral on Curve C (IR3)

[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! :)

 

[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.

 

[portuguese]

But it is still correct?

 

[haruspex]

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

 

[portuguese]

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

 

[HallsofIvy]

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?

 

[haruspex]

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?

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

 

[HallsofIvy]

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

 

[portuguese]

Yeah, my answer is 0

 

[HallsofIvy]

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.

 

[portuguese]

I found my mistake. I wrote:

The integral becomes
\int_0^{2\pi} sin(\theta)(-\sqr{2}sin(\theta)d\theta)<span style="font-size: 22px">+ (\sqr{2}cos(\theta)+ sin(\theta))cos(\theta)d\theta</span>
=...<br />

<br /> <br /> It&#039;s not 0, you&#039;re right. Thanks a lot! ;)

 

[haruspex]

I found my mistake.

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 canceled 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.