Calculate surface integral $\iint_S \nabla \times \mathbf{F} \bullet \mathbf{n} \; dS$ with stokes, when $$\mathbf{F}=\left\langle\frac{5y(z-1)}{6},xz, 6e^{xy}\cos{z}\right\rangle$$ and $S$ is surface $x^2+y^2+z^2=4$, $z \geq0$, which's positive side is the above side.

My work so far.. $$S=\{r\in[0,2], \phi \in [0,2\pi]\}$$ $$\begin{align} \int_0^{2\pi}\int_0^2 ((\frac{\overbrace{z}^{=0}+5}{6})\mathbf{k}\bullet \mathbf{k}) r \; d rd \phi &= \frac 56 \times \text{area of circle with radius of 2} \\ &=\frac 56\times4\pi = \frac {10}{3} \pi \end{align}$$

  • $\begingroup$ For the line integral you need the boundary curve $C$ of $S$. $\endgroup$ – Ted Shifrin Apr 6 '14 at 15:46

Hint: The flux of $\text{curl}\,\mathbf F$ is the same across any oriented surface with the same boundary curve.

  • $\begingroup$ How can I find the boundary curve for the surface? Is it the parameterization in my edit. $\endgroup$ – ELEC Apr 6 '14 at 17:42
  • $\begingroup$ The boundary curve is the circle $x^2+y^2=4$, $z=0$. You restrict your parametrization to $\phi=\pi/2$, $0\le\theta\le 2\pi$. My suggestion, however, was to fill this circle in and integrate over the region it bounds in the $xy$-plane. $\endgroup$ – Ted Shifrin Apr 6 '14 at 17:50
  • $\begingroup$ So it would be that $\mathbf{r}=\langle 2\cos\phi, 2\sin\phi\rangle$ and then evaluate $\mathbf{F}(\mathbf{r}(t))$ and $\mathbf{r}'$ and just do the integral $\oint_S \mathbf{F} \bullet d\mathbf{r}$. $\endgroup$ – ELEC Apr 6 '14 at 19:00
  • $\begingroup$ No, use $\theta$ here, not $\phi$. And I'm still recommending you avoid the line integral altogether. $\endgroup$ – Ted Shifrin Apr 6 '14 at 19:02
  • $\begingroup$ Okey right. I'm still not sure what the integral would look like you suggest I should use. The one in $S$ with the normal and all that? $\endgroup$ – ELEC Apr 6 '14 at 19:10

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