Application of Stoke's theorem on a swimmer in a swimming pool

Question

Problem says:

A circular swimming pool has a small, circular island in the middle. There is a current with components, in Cartesian components

$$\vec{v}= \frac{1}{x^2+y^2}(-y,x)$$

To a swimmer who swims around the pool this vector represents a force that he has to work against. The curl of this vector field is zero. (You need not prove that.) So a naive application of Stokes' theorem says that swimming around the pool (ending at the point of departure), requires no work. Carefully, study this conclusion in order to understand why it is false. Write an argument.

Answer 1

We will compute the work done going against the current on a circle $C$ of radius $1$. Note that in order to go against the current, we need to swim clockwise.

Let $\gamma(t) = (\cos(t),\sin(t)),\ \gamma'(t) = (-\sin(t),\cos(t)), \ t\in[0,2\pi[$ $$ \oint_C \vec{v} \cdot d\vec{l} = \int_0^{2\pi} \frac{1}{\cos(t)^2 + \sin(t)^2} (-\sin(t),\cos(t)) \cdot (-\sin(t),\cos(t)) dt = \int_0^{2\pi} 1 dt = 2\pi $$

The fact that the $ curl = 0 \implies \oint F dl = 0$ works under the condition the domain of $F$ is convex inside the loop.

Answer 2

You can't directly apply stokes theorem here.

If $ C_{sw} $ is the curve the swimmer follows, and $C_{is} $ is the boundary of the island (both traversed anti-clockwise), You will have to apply stokes theorem to the region in between these two curves. This will result in the difference between line integrals along $C_{sw}~ and ~C_{is} $ This is why the naive application of stokes theorem is invalid.

Have a look at this pdf from MIT ocw, it addresses a very similar question. (The theorem is referred to as 'Green's theorem' here)

https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-c-greens-theorem/session-71-extended-greens-theorem-boundaries-with-multiple-pieces/MIT18_02SC_we_71_comb.pdf