# Finding the flux of $\iint \vec F\hat n\;ds$

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

$$\iiint\bigg(4-2y+2z\bigg)dxdydz$$

Is it correct so far?

• 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 ? – Svetoslav Jul 29 '15 at 17:22
• If so, then you do not need Stokes. This is just a surface integral of the second kind. – Svetoslav Jul 29 '15 at 17:23
• 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 $\hat n \times \vec F$? – Mark Viola Jul 29 '15 at 17:34
• Just FYI ... The Curl maps a vector into a vector. Your result is a scalar. – Mark Viola Jul 29 '15 at 17:41

## 2 Answers

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 $0 \leq z \leq 3$.

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

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

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

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

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

$$\iiint_{V}\bigg(4-4y+2z\bigg)dxdydz=\int_0^3dz\int_0^2d\rho\int_0^{2\pi}\rho(4-4\rho\sin\varphi+2z)d\varphi$$

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