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

Question

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?

Answer 1

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

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

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

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

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

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

Answer 2

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 $0 \leq z \leq 3$.