I need to find the flux $\displaystyle\iint \vec F\hat n\;ds$ of the vector feild $\vec F=4x \hat i-2y^2\hat j+z^2 \hat k$ throughe the surface $S=\{(x,y,z):x^2+y^2=4,z=0,z=3\}$

My attempt:

(I'm not sure that I know what I'm doing)

curl $\vec F=\nabla\times F=\bigg(\frac{\partial(4x)}{\partial x}+\frac{\partial (-y^2)}{\partial y}+\frac{\partial (z^2)}{\partial z}\bigg)=\bigg(4-2y+2z\bigg)$

Now applying Stock's theorem


Is it correct so far?

  • $\begingroup$ Do you mean that the surface is closed, i.e the cylinder with the one base at z=0 and the other at z=3 ? $\endgroup$ – Svetoslav Jul 29 '15 at 17:22
  • $\begingroup$ If so, then you do not need Stokes. This is just a surface integral of the second kind. $\endgroup$ – Svetoslav Jul 29 '15 at 17:23
  • $\begingroup$ Is that a tensor $\vec F\,\hat n$ in the surface integral or is there intended to be an inner product $\hat n \cdot \vec F$ or perhaps a cross product $\hat n \times \vec F$? $\endgroup$ – Mark Viola Jul 29 '15 at 17:34
  • $\begingroup$ Just FYI ... The Curl maps a vector into a vector. Your result is a scalar. $\endgroup$ – Mark Viola Jul 29 '15 at 17:41

You must use the Gauss Theorem (divergent theorem), since it deals with flux:

$$ \nabla F = 4-4y+2z $$

$$ \int_{V} \nabla F dV = \int\int \vec{F}\vec{n}dS $$

Your volume is define by the cylinder $x^2 + y^2 = 4$ and $0 \leq z \leq 3$.

  • $\begingroup$ You have a vector on the left-hand side and a tensor on the right-hand side. $\endgroup$ – Mark Viola Jul 29 '15 at 17:35
  • $\begingroup$ I guess that the inner product $\vec{F} \cdot \vec{n}dS$ is implicit in his problem. $\endgroup$ – Gabs Jul 29 '15 at 17:39
  • $\begingroup$ Is it? Why does one assume that? It could as easily be $\vec F \times \hat n$, could it not? $\endgroup$ – Mark Viola Jul 29 '15 at 17:41
  • $\begingroup$ He needs to determine the flux... Stock's theorem does not evaluate fluxes, but potentials over closed loops. $\endgroup$ – Gabs Jul 29 '15 at 17:42
  • $\begingroup$ Plus, in some text books, the dot product is assumed when the dot is omitted. $\endgroup$ – Gabs Jul 29 '15 at 17:43

div$\vec F=\bigg(\frac{\partial(4x)}{\partial x}+\frac{\partial (-y^2)}{\partial y}+\frac{\partial (z^2)}{\partial z}\bigg)=\bigg(4-2y+2z\bigg)$

$$\iint_{S}\vec F\cdot \hat n \;ds=\iint_{V}div\vec F \;dv=\iiint_{V}\bigg(4-2y+2z\bigg) \;dxdydz$$

$$x=\rho \cos \varphi,\; y=\rho \sin \varphi, \; z=z$$

$$0\leq z \leq3, \; 0\leq \rho \leq 2, \; 0\leq\varphi\leq 2 \pi$$


$$=\int_0^3dz\int_0^22\pi\bigg(4\rho+2\rho z\bigg)d \rho=48\pi+36\pi=\boxed{\color{red}{84\pi}}$$


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