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Im struggling to solve this problem :

Consider F$(x,y,z) = (y,z,x)$ and the surface $\phi(u,v) = (u,v,sin(u^2 + v^2))$ where $(u,v)$ belongs to some disc centred at $(0,0)$ and $r=3$. Find the surface integral using Stoke's Theorem.

So far I have parametrised the curve so that $$ \gamma(t) = (3cos(t),3sin(t), 0)$$

I'm slightly unsure of the method used to solve this integral using Stoke's Theorem. What should i be taking the curl of?

The correct answer should be zero which is not what im getting.

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    Does "the surface integral" refer to the integral of the _curl_ of $F$ over the graph provided, or to the integral of $F$ itself? – 2017-01-12
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    It asks for $\int_S (\nabla \times F) * n ~d\sigma$ – 2017-01-12
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    Your curve is not quite OK (the $z$-coordinate should be $\sin(9) \neq 0$), but even when that change is made, the boundary integral $\int_{\gamma} F \cdot d\mathbf{s}$ (which Stokes's theorem guarantees is equal to the surface integral you gave) is not zero. Not sure what to suggest other than "recheck the problem statement, and that the answer matches the problem number".... – 2017-01-12
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    It's definitely the corresponding answer for the given question. Why should the $z$-coordinate be $sin(9)$ ? Im sure the parametrisation of a disc is $x=rcos(t), y=rsin(t), z = 0$ ? – 2017-01-12
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    You're right about the disk, but your parametrization needs to cover the surface $S$, which is part of the graph $z = \sin(x^{2} + y^{2})$. – 2017-01-12

1 Answers 1

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Hint: You can write $$F=\nabla\times G$$ where $$G(x,y,z)=\left (\frac{z^2}{2},\frac{x^2}{2},\frac{y^2}{2}\right )$$

in order to use Stokes' theorem to express $$\int_S F\cdot \mathrm d\Sigma=\int_S \left (\nabla \times G\right ) \cdot \mathrm d \Sigma=\oint_{\partial S}G\cdot \mathrm d s$$