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Let $\Sigma$ be the portion of the paraboloid $z=1-x^{2}-y^{2}$ where $x\geq0$ and $z\geq0$. Consider the vector fields $\mathbf{u}=(xy,xz^{2},x^{2}y)$ and $\mathbf{v}=\mathrm{curl}\,{\mathbf{u}}$. Compute the flux of $\mathbf{v}$ through $\Sigma$.

The solution manual solves the problem by applying Stokes theorem as follows-

$$\iint_\Sigma \nabla \times \mathbf{u} \cdot d\boldsymbol{\Sigma} = \oint_{\partial\Sigma_1} \mathbf{u} \cdot d\mathbf{r}\ +\oint_{\partial\Sigma_2} \mathbf{u} \cdot d\mathbf{r}\,$$

where $\partial\Sigma_1$ and $\partial\Sigma_2$ are the boundary curves of $\Sigma$. $\partial\Sigma_1$ in the xy-plane and $\partial\Sigma_2$ in the yz-plane.

The problem for me is while I understand that $\mathbf{v}=\mathrm{curl}\,{\mathbf{u}}$ I feel that to apply Stokes correctly one should instead calculate

$$\iint_\Sigma \nabla \times (\nabla \times \mathbf{u}) \cdot d\boldsymbol{\Sigma} = \oint_{\partial\Sigma_1} \mathbf{v} \cdot d\mathbf{r}\ +\oint_{\partial\Sigma_2} \mathbf{v} \cdot d\mathbf{r}\,.$$

Why am I wrong?

Any help appreciated.

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  • $\begingroup$ They asked to find the flux of ${v}$ through $\Sigma$, not $\text{curl }{v}$. $\endgroup$ – dromastyx Aug 10 '17 at 11:16
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Both formulas you wrote are correct, but the first one is the flux of $\mathbf{v}$ through $\Sigma$: $$\text{flux}=\int \int_\Sigma\mathbf{v} \cdot d\mathbf{\Sigma} =\int \int_\Sigma \nabla \times \mathbf{u} \cdot d\mathbf{\Sigma}$$ while the second one is the flux through $\Sigma$ of $\nabla \times \mathbf{v}$

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