help with stokes theorem problem

Question

Suppose $F=(5x−5y)\hat{i}+(x+2y)\hat{j}$. Using Stokes' Theorem

Find the circulation of $F$ around the circle $C$ of radius $7$ centered at the origin in the $yz$-plane, oriented clockwise as viewed from the positive $x$-axis.

Answer 1

Your curl is correct but your final result is not. Even though you seem to understand what you're doing I will give full detail for my own benefit and the benefit of future readers:

Let $F: \mathbb R^3 \to \mathbb R^3$ be given by $F(x,y,z) = (5x -5y) \textbf{i} + (x + 2y)\textbf{j}$.

$F$ is the flow velocity of your liquid. By definition, the circulation is given by

$$ \oint_C F \cdot dl$$

And by Stokes' theorem, if we let $C=\partial S$ be the boundary curve of some area $S$ then

$$ \oint_C F \cdot dl = \int \int \nabla \times F \cdot \textbf{n} dS$$

where $\textbf{n}$ is the normal to the surface we're integrating over. So in our case, $\textbf{n}=\textbf{i}$.

$\nabla \times F$ is given by

$$ \left( {\partial F_z \over \partial y}- {\partial F_y \over \partial z} \right) \textbf{i} + \left( {\partial F_x \over \partial z}- {\partial F_z \over \partial x} \right)\textbf{j } + \left( {\partial F_y \over \partial x}- {\partial F_x \over \partial y} \right)\textbf{k} = (0-0)\textbf{i} + (0-0)\textbf{j} + (1+5)\textbf{k} = 6 \textbf{k}$$

So that

$$ \oint_C F \cdot dl = \int \int \nabla \times F \cdot \textbf{n} dS = \int_0^7 \int_0^{2\pi} 6 \textbf{k} \cdot \textbf{i} dr d\phi = \int_0^7 \int_0^{2\pi}0 dr d\phi=0$$

To be sure let's see if we get the same result by direct calculation (without using Stokes):

$C = \gamma(t) = (0, 7\cos t, 7\sin t)$

$\gamma'(t) = (0, -7\sin t, 7 \cos t))$

$F(x,y,z) = (5x -5y, x + 2y, 0)$.

Then

$$ \oint_C F dl = \int_0^{2\pi} f(\gamma(t)) \cdot \gamma'(t) dt = -98 \int_0^{2\pi} \sin t \cos t dt = 0$$