Find the outward flux of the vector field $F=(x^3,y^3,z^2)$ across the surface of the region that is enclosed by the circular cylinder $x^2+y^2=49$ and the planes $z=0$ and $z=2$.

  • $\begingroup$ What exactly is your problem? Do you know the how flux is calculated? Do you know what flux is? :) Show us your attempt. $\endgroup$ – u_sre Jul 4 '19 at 16:01

You can use the divergence theorem to evaluate the outward flux of the vector field.

Divergence theorem states the following:

Divergence Theorem

In other words we can simply add up the divergence in the region bound by our surface $S$, in order to calculate the outward flux of our vector field across our surface $S$.

You can read more here: https://en.wikipedia.org/wiki/Divergence_theorem

Now we can apply divergence theorem:

$$\nabla\cdot\mathbf{F}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}$$


Now we need to simply integrate over our region so we can evaluate: $$\int_0^2 \int_{-7}^7 \int_{-\sqrt{49-y^2}}^{\sqrt{49-y^2}} 3x^2+3y^2+2z \hspace{1mm} dx dy dz$$ But our integral is much easier if we use polar coordinates: $$\int_0^2 \int_{0}^{2 \pi} \int_{0}^{7} (3r^2+2z)r \hspace{1mm} dr d \theta dz$$

Evaluating this integral should get you $7339\pi$

  • $\begingroup$ i got the same answer but its not correct $\endgroup$ – Ashish Paliwal Jul 4 '19 at 16:40
  • $\begingroup$ I checked the answer with a calculator it should be correct. Did you copy down the question correctly? Are you sure the last term was $z^2$ and not $z^3$ $\endgroup$ – Anirudh Jul 4 '19 at 16:49

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