0
$\begingroup$

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 $z=2$.

$\endgroup$
1
  • $\begingroup$ 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
0
$\begingroup$

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:

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

$$\nabla\cdot\mathbf{F}=3x^2+3y^2+2z$$

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 $7339\pi$

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.