Compute the circulation of the vector field?

Question

Compute the circulation of the vector field H=<-y^3, x^3> over the boundary of the region D={x^2+y^2<=1, y>=0}.

Answer 1

Hello Sun,

I realize that I come to this question likely long past its expiration date, so to speak, but I noticed it and felt I had to comment. 

The answer above, left by Roman C., is beautifully laid out, but I think Roman made one minor computational error in computing the differential ds, which subsequently has caused an incorrect answer.  I have no doubt that this was merely an oversight, and I do not comment here in order to raise any trouble.

We can, incidentally, see why the answer cannot be zero, by resorting to Stokes' Theorem, which asserts that the circulation about C,

∫_C v · T ds = ∫∫_R curl v dA,

for continuously differentiable vector field v = p(x,y)i + q(x,y)j and C positively oriented. 

Recall that

curl v = -(∂p/∂y) + (∂q/∂x) = 3(x² + y²)

in this case, which is not zero on R (the interior of C). 

To find the circulation, simply evaluate the integral by Stokes' Theorem:

∫_C v · T ds = ∫∫_R curl v dA = 3∫₀¹ ∫₀^π r² (r dθ dr) = 3π/4.

The most important aspect of this problem (in my opinion) is that we are able to use Stokes' Theorem to convert a line integral into an easy (in polar coordinates) integral over a region.

Regards,

Hassan H.

Answer 2

D is a semicircle. ∂D can be piecewise parametrized as s(t) = <x(t), y(t)>.

s(t) = <cos t, sin t> for 0 ≤ t < π

s(t) = <cos t, 0> for π ≤ t < 2π

This will trace the path exactly once.

we need the differential ds.

ds = <-sin t, -cos t>dt for 0 ≤ t < π

ds = <-sin t, 0>dt for π ≤ t < 2π

Now we need to express H parametrically. From the above, we get the following.

H = <-sin³ t, cos³ t> for 0 ≤ t < π

H = <0, cos³ t> for π ≤ t < 2π

You can get the circulation as follows.

∫_∂D H·ds

= ∫₀^π <-sin³ t, cos³ t>·<-sin t, -cos t> dt + ∫_π^2π <0, cos³ t>·<cos t, 0> dt

= ∫₀^π (sin⁴ t - cos⁴ t) dt + ∫_π^2π 0 dt

= ∫₀^π (sin² t + cos² t)(sin² t - cos² t) dt

= ∫₀^π 1*(-cos 2t) dt

= (-1/2)sin 2t |₀^π

= (sin 2π - sin 0)/2

= 0