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Verify Stokes's Formula for $\textbf{F}(x,y,z)=(3y,-xz,yz^2)$, where $S$ is the surface of the paraboloid $2z=x^2+y^2$ bounded by the plane $z=2$.

So I need to compute the integral using the formula $\iint_S \text{curl}~\textbf{F} \dot ~\eta ~d\sigma$ and the formula $\int_\Gamma \textbf{F}~\dot~\textbf{T}~ds$ and get the same answer.

I have a formula that says $$\int_\Gamma \textbf{F} \cdot \textbf{T}~ds = \int_a^b \textbf{F}(\gamma(t)) \cdot \gamma'(t)dt,$$ so I thought that the line that I am integrating across is $4=x^2+y^2$, so I said $\gamma(t)=(2\text{cos}(t),2\text{sin}(t),2)$, with $0 \leq t \leq 2\pi$. From here it's a pretty straight forward computation, but I'm not sure I set it up correctly. Does this look right?

Now for the first integral, I have $$\iint\limits_S \text{curl}~\textbf{F} \cdot \eta ~d\sigma = \iint\limits_R \text{curl}~\textbf{F}(\textbf{r}(u,v)) \cdot \frac{\partial \textbf{r}}{\partial u}\times\frac{\partial \textbf{r}}{\partial v} du \, dv.$$ However, I can't figure out what $\textbf{r}(u,v)$ should be. I tried using polar coordinates but it was pretty ugly. Any tips?

user126004
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1 Answers1

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Your setup for the line integral is correct. For the second, there are some ways to parametrize. I'm going to show two. $$\text{First: } \begin{align} x & = u, \\ y & = v, \\ z & = \frac{u^2+v^2}{2}, \end{align} \quad -2 \leq u,v \leq 2, \; u^2+v^2 \leq 4.$$ Notice this is the parametrization of any graph. $$\text{Second: } \begin{align} x & = r \cos \theta, \\ y & = r \sin \theta, \\ z & = \frac{r^2}{2}, \end{align} \quad 0 \leq r \leq 2, \; 0 \leq \theta \leq 2 \pi.$$ This is in terms of polar coordinates. I recommend the second.

Mark Fantini
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  • So I computed the line integral and got $-20 \pi$. But if I do that parametrization and say $\alpha(r,\theta)=(rcos(\theta),rsin(\theta),\frac{r^2}{2})$, I get $\frac{\partial \alpha}{\partial r}\times\frac{\partial \alpha}{\partial \theta} = (-r^2cos(\theta),-r^2sin(\theta),r)$, not $(0,0,1)$. Did I just compute this wrong or am I misunderstanding what you are saying? – user126004 Apr 14 '14 at 16:28
  • @user126004 My comment was irrelevant, you are correct. It was valid if you took the disk $x^2+y^2=4$ at $z=2$, but that's not the surface you are integrating. Sorry. – Mark Fantini Apr 15 '14 at 03:58