Given the velocity components for a flow
$$ u = 16x^2+y, \hspace{10pt} v = 10, \hspace{10pt} w = yz^2 $$
and a rectangular region $R$ in the $xy$-plane formed by the points $(0,0)$, $(10,0)$, $(10,5)$, $(0,5)$, determine the circulation for this flow field around the border of $R$ (traveling counter-clockwise).
I have that the circulation is
$$ \displaystyle\Gamma=\oint_R \textbf{u} \cdot d\textbf{x} $$
where $ \textbf{u}=\langle u, v, w \rangle$
I broke up $R$ into four paths with the following parameterization for $0\le t \le 1$:
- $C_1 : (0,0)\rightarrow(10,0)$ $\hspace{1in}$ $\textbf{r}_1 = \langle 10t,0,0 \rangle$
- $C_2 : (10,0)\rightarrow(10,5)$ $\hspace{.91in}$ $\textbf{r}_2 = \langle 10,5t,0 \rangle$
- $C_3 : (10,5)\rightarrow(0,5)$ $\hspace{1in}$ $\textbf{r}_3 = \langle 10-10t,5,0 \rangle$
- $C_4 : (0,5)\rightarrow(0,0)$ $\hspace{1.09in}$ $\textbf{r}_4 = \langle 0,5-5t,0 \rangle$
I then calculated each line integral over its path with its specific parameterization in order to obtain the circulation. However, when I did this, I found the circulation to be
$$ \displaystyle\Gamma=\int_{t=0}^1 16000t^2dt + \int_{t=0}^1 50dt -\int_{t=0}^1 16000(t-1)^2dt - \int_{t=0}^1 50dt = 0$$
Am I doing something wrong? Does this just mean that the flow is irrotational?
One way to solve this problem is using Stokes' theorem: $$ \oint \vec{F}\cdot \vec{dl} = \iint \operatorname{curl}(F)\cdot \hat{n} dS $$ In this case the whole surface is in the $xy$-plane so $\hat{n} = (0,0,1)$ and stokes' theorem becomes Green's theorem: $$ \oint udx + vdy = \iint \left (\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right ) dxdy$$ Substituting $u=16x^2+y, v=10$ we get that the circulation is $$ \int_0^5 \int_0^{10} (0-1)dxdy=-50 $$
Your way is correct too, but I think you have a minor problem with your calculation:
I don't understand your parameterization and your substitution. For the first part of the curve I would use $\gamma_1(t)=(t,0) , 0\leq t \leq 10 $ and get $ \int_0^{10} 16t^2 dt $ and for the third part $\gamma_3 (t) = (10-t,5) , 0\leq t\leq 10 $ and get $$ \int_0^{10} -(16(10-t)^2 + 5)dt $$ when you add them you will indeed get $-50$.