2
$\begingroup$

I am lost with that problem, and I cannot continue.

The problem asks to calculate that:

$$\oint_S\ \vec F\cdot \vec {dS}$$ Being a vector defined by:

$$F = x^2\hat a_x+ y^2\hat a_y + (z^2-1)\hat a_z$$

and S is defined by these Cylindrical coordenates: $$r = 2; 0<z<2; 0\leqΦ\leq2π$$

I converted F to cylindrical coordenates for having both in the same system. I found for cylindrical system, derivative area is given by: $$dS = (r.dΦ.dz)\hat ar + (dr.dz)\hat aΦ + (r.dr.dΦ) \hat az$$

But is that the best way to do that with surface integrals? Seems, that integral gives a lot of job. Please help me.

$\endgroup$
  • 1
    $\begingroup$ What exactly are these expressions $\hat ar, \hat a\Phi, \hat az$ at the end? (By the way - you should go check the wording of your problem carefully to determine if $S$ is eactly what you've discussed here, or should it also include the dlsks at $z = 0$ and $z=2$ that would close off the surface. If so, then you have two more integrals to do.) $\endgroup$ – Paul Sinclair Mar 30 at 0:49
  • $\begingroup$ And a hint: Note that $r$ is constant on $S$. $\endgroup$ – Paul Sinclair Mar 30 at 0:51
  • $\begingroup$ These expression $\hat ar,\hat aΦ,\hat az$ are derivation of the surface area of a cylinder, written in Cylindrical Coordinates. But I do not need to use in $Φ$ direction, once I need to calculate area of top, bottom and side. $\endgroup$ – user2793412 Mar 30 at 3:54
  • $\begingroup$ Are you familiar with differential forms? $\endgroup$ – Bertrand Wittgenstein's Ghost Mar 30 at 4:52
  • $\begingroup$ That doesn't answer my question. Note the "exactly" part. How are you defining these? They look like you are multiplying the same vector by 3 coordinates, which makes no sense at all. $\endgroup$ – Paul Sinclair Mar 30 at 14:14
0
$\begingroup$

Hint

By divergence theorem, we can write$$\oint_S \vec F\cdot \vec{dS}=\iiint_V \nabla\cdot fdV$$where$$\nabla\cdot f{=2x+2y+2z\\=2r(\sin\phi+\cos \phi)+2z}$$

$\endgroup$

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.