# Vector Line Integral

1. Dec 9, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
Integral closed line Integral ∫F ds where F = <y+sin(x^2), x^2 + e^y^2> and C is the circle of radius 4 centered at origin.

2. Relevant equations

3. The attempt at a solution
so ds = c'(t)dt I believe...
where
c(t) = <4cos(t),4sin(t)>
c'(t) = <-4sin(t),4cos(t)>

Then we have
F = <y+sin^2(x),x^2+e^y^2>
where
y = 4sin(t)
x = 4cos(t)

F = <4sin(t)+sin^2(4cos(t)),(4cos(t))^2+e^(4sin(t))^2>
F dot c'(t) = (-16sin^2(t) - 4sin(t)*sin^2(4cos(t))) + (64(cos(t)^3)) +(e^(4sin(t))^2)*4cos(t)

and then take the integral of that with respect to dt from 0 to 2pi?

2. Dec 9, 2013

### CAF123

In principle yes, but the integrand is a mess. Have you covered anything in your course that relates line integrals to double integrals?

3. Dec 9, 2013

### zahbaz

Looks correct to me. The integral isn't as evil as it appears. I tried to integrate the four final terms separately... For three of the terms, it seems that with a bit of w-substitution, you should be able to redefine the bounds to a form where you won't need to evaluate the integral at all. An integral from [a,a] is zero after all :)

4. Dec 9, 2013

### PsychonautQQ

Do you mean Stokes Theorem? If I turn this integral into ∫∫∇xF dS then what do I use for dS? does it become n(dudv) ?? where n is the normal vector? if so how do I find the normal vector of a circle of radius 4 centered at the origin but it doesn't tell you what plane the circle lies in?

5. Dec 10, 2013

### CAF123

Yes, and since the surface in question is planar, we can use Green's theorem, the special case of Stokes.

The normal vector will not depend on what plane the circle lies. Your analysis in the OP assumes a circle in the xy plane. dS is computed as usual: $$d\underline{S} = \hat{n}dS = |\underline{r}_r \times \underline{r}_{\theta}| dA\, \hat{n}$$

You will see that the double integral obtained from computing the dot product of the surface normal and the curl of F is the same double integral you would have obtained by directly using Green's theorem.

6. Dec 10, 2013

### PsychonautQQ

so I take (del x F) even though there is no z (k) component? I haven't done a cross product with only 2 variables before

7. Dec 10, 2013

### CAF123

There is no notion of a cross product for vectors in R2. In your case, take the z component of F to be zero.

8. Dec 10, 2013

### PsychonautQQ

Okay cool, so I took ∇xF and got (2x+1)k

Now I have to take
∫∫(∇xF) dS

I always run into the problem of trying to figure out what dS is...

dS = (normal vector)dxdy? as in the vector normal to the circle? Wouldn't that be (0,0,-1) or (0,0,1) ?

9. Dec 10, 2013

### CAF123

Is there a sign error there?
Yes, the normal vector $\hat{n} = \hat{k}$ for a positive orientation of the curve. What is the transformation of the area element dxdy to polar coordinates?

10. Dec 10, 2013

### PsychonautQQ

dxdy turns to 4drd(theta)

11. Dec 10, 2013

### PsychonautQQ

you are right,,
∇xF = (2x-1)k

∫∫∇xF dS
where dS = <0,0,1> r drdθ?
is that correct?

with
x = rcosθ the integral becomes
∫∫(2cos(θ)r^2 - r) drdθ
where dr goes from 0 to 4
and θ goes from 0 to 2pi

12. Dec 10, 2013

### CAF123

That is all correct. If you have time, you can confirm you get the exact same integral using Green's theorem.

13. Dec 10, 2013

### PsychonautQQ

oh wow greens theorem duh.
Are you getting the answer -16pi?

14. Dec 10, 2013

### zahbaz

That's what I got from the non-Stokes approach:

∫F dot c'(t) dt from [0,2pi] = -16pi

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Not sure if you've used Wolfram Alpha before, but it serves as a great check on your integration. This particular integral was a bit, erm, long for the free service, but the answer -50.26 matches -16pi

Wolfram input of your original integral