# Vector Line Integral

## Homework Statement

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.

## 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?

CAF123
Gold Member
In principle yes, but the integrand is a mess. Have you covered anything in your course that relates line integrals to double integrals?

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 :)

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

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?

CAF123
Gold Member
Do you mean Stokes Theorem?
Yes, and since the surface in question is planar, we can use Green's theorem, the special case of Stokes.

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?
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.

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

CAF123
Gold Member
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
There is no notion of a cross product for vectors in R2. In your case, take the z component of F to be zero.

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) ?

CAF123
Gold Member
Okay cool, so I took ∇xF and got (2x+1)k
Is there a sign error there?
dS = (normal vector)dxdy? as in the vector normal to the circle? Wouldn't that be (0,0,-1) or (0,0,1) ?
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?

dxdy turns to 4drd(theta)

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

CAF123
Gold Member
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
That is all correct. If you have time, you can confirm you get the exact same integral using Green's theorem.

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

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