I'm having some trouble calculating vector fields through surfaces. After attempting a few and being dissapointed with a wrong answer multiple times I figured I must be doing something wrong in the process. I think its the parametrization I'm struggling with.

Oftentimes there's a surface I can easily parametrize in Cartesian coordinates (sometimes partially) which can also be written using cylindrical or spherical coordinates. It's this part that gives me trouble, I don't really how to combine a change of coordinates and a parametrization.

I looked around at a few examples online and tried applying what I learnt there without much positive result, here's one of the attempts:

Integrate the function $\vec F(x,y,z) = (2x,-3y,z)$ through the surface $\Sigma$ defined as the part of the cylinder $x^2+y^2=1$ between $z=0$ and $z=x+2$.

Parametrization $\vec\phi$:
$x = \cos\theta$
$y = \sin\theta$
$z = z$
$\frac{\partial\phi}{\partial z}\times\frac{\partial\phi}{\partial\theta}=(-\cos\theta,-\sin\theta,0)$

Apply the formula: $\iint_\Sigma\vec F(x,y,z) .d\vec\sigma = \iint_K\vec F(\vec\phi(z,\theta)).(\frac{\partial\vec\phi}{\partial z}\times\frac{\partial\vec\phi}{\partial\theta}(z,\theta))$

Which gives $\int_0^{2\pi}d\theta\int_0^{\cos\theta+2}(2\cos\theta,-3\sin\theta,z).(-\cos\theta,-\sin\theta,0)dz$

Solving this integral doesn't give me the correct result, so I must've made an error along the way but I just can't find it. Again, this is just an example, it's this type of integrating that's giving me a hard time.

Here's the further solution:

$= 0 + 0 + (-4\pi) + 6\pi = 2\pi$

The correct answer according to my textbook is $-2\pi$

  • $\begingroup$ What is the answer that you can't seem to get? The last equation, it should be written as $\int_0^{2\pi}\int_0^{2+\cos(\theta)}(2\cos(\theta), -3\sin(\theta),z)\cdot (-\cos(\theta), -\sin(\theta),0)dz d\theta$. You can't split them up like that. $\endgroup$
    – MathNewbie
    Commented May 30, 2015 at 20:46
  • $\begingroup$ @MathNewbie: I think it is not split, it is a integral into another, see en.wikipedia.org/wiki/Multiple_integral#Normal_domains_on_R2 $\endgroup$ Commented May 30, 2015 at 21:10
  • $\begingroup$ @enzotib, ah I see. Looking at it, never seen it done that way. I just find it quite odd. $\endgroup$
    – MathNewbie
    Commented May 30, 2015 at 21:15
  • $\begingroup$ I've added the rest of the solution. $\endgroup$
    – Joshua
    Commented May 31, 2015 at 12:11
  • 1
    $\begingroup$ What orientation should your surface have? You didn't specify one. $\endgroup$ Commented May 31, 2015 at 12:48


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