# Finding the flux of a cylinder using the Divergence Theorem

I have to calculate the flux of the $$F(x,y,z)=(x+\ln(yz^{2}),y+e^{-(x^{2}+z^{2})},2z)$$ through the cylinder (without its bases): $$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$$ for $z\in[0,4]$.

I know that if $S_{1}, S_{2}$ are the cylinder bases and $S_{3}$ its "body", then let $$S=S_{1}\cup S_{2}\cup S_{3}$$ so I can apply the divergence theorem in $S$. When I do it, I get: $$\iint \limits_{S} F\;dS=96\pi$$ But I know that $$\iint \limits_{S} F\;dS=\iint \limits_{S_{1}} F\;dS+\iint \limits_{S_{2}} F\;dS+\iint \limits_{S_{3}} F\;dS$$ So, to finish my exercise, I have to evaluate $$\iint \limits_{S_{1}} F\;dS\quad\text{and}\quad\iint \limits_{S_{2}} F\;dS$$ Let $S_{1}$ be parameterized by $$\sigma(r,\theta)=(2r\cos\theta,3r\sin\theta,4)$$ When I use this parameterization to calculate $\iint \limits_{S_{1}} F\;dS$, I get $$\sigma_{r}\times\sigma_{\theta}=(0,0,6)$$ If I use this vector above in the formula of the flux through a surface, I get: $$\iint \limits_{S_{1}} F\;dS=288\pi$$ My question is: my teacher used this parameterization for $S_{1}$: $$\sigma(u,v)=(u,v,4)$$ and then he got $$\iint \limits_{S_{1}} F\;dS=48\pi$$ So, which parameterization should I use? Could he do like that? Just take $(u,v,4)$?

## 1 Answer

Note that the flux component along $z$ is equal to $0$ for $z=0$ and it is equal to $8$ for $z=4$. Thus the overal flux through the bases is

$$8\cdot A=48 \pi$$

Note that with your first parameterization we obtain the same result, that is

$$\iint \limits_{S_{1}} F\;dS=\int_0^{2\pi}d\theta\int_0^1 (F_x,F_y,8)\cdot (0,0,6r)\,dr=2\pi\cdot 48\cdot[r^2/2]_0^1=48\pi$$

• But shouldn't be $dS$ just $dr d\theta$? – mvfs314 Jun 29 '18 at 19:15
• We have $$\sigma_{r}\times\sigma_{\theta}=(0,0,6r)$$ – user Jun 29 '18 at 19:28