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I need to integrate $\vec F(x,y,z) = (2x,-3y,z)$ through a surface $\Sigma$.
$\Sigma$ is defined as the surface of the cylinder $x^2+y^2=1$ (so without the top and the bottom) that is confined between the plane $z=0$ and $z=x+2$.

I think I can't project it to my x-y plane since that didn't really work out when I tried it, so then I tried splitting it in 2 along the x-axis and then projecting it to the z-x plane. Then I split the projection in 2 again so I get a bottom and a top half, resp. a rectangle and a triangle.

Now lets say I take the top half (triangle) of the part where y is positive.

My parametric representation would be:
$x=u$
$y=\sqrt{1-u^2}$
$z=v$

My boundaries would be:
$u=-1 \rightarrow u=+1$
$v=1 \rightarrow v=x+2$

Now I know everything to complete the formula: $\iint_\Sigma\vec F.d\vec\sigma = \iint_K\vec F(\vec\phi(u,v)).(\frac{\partial\vec\phi}{\partial u}\times\frac{\partial\vec\phi}{\partial v})(u,v)du dv$

The cross product gives me $(\frac{-u}{\sqrt{1-u^2}};1;0)$ and so I get the following:
$\iint_K\frac{-2u^2}{\sqrt{1-u^2}}-3\sqrt{1-u^2} dudv$

I'm unsure how to proceed from here or if what I did was even correct... My issues are:

  • I don't know how to solve the integral, I know I can split it up in the integral of an integral and then integrate one to $v$ and the other to $u$ (with the appropriate boundaries I defined earlier) but I'm unsure how to actually solve it.
  • How will I add these 4 parts up? Do I just add them or is there more to it?
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  • $\begingroup$ Might I respectfully suggest that you try parametrizing $\Sigma$ in cylindrical coordinates $(\theta,z)$? $\endgroup$ – Ted Shifrin Jun 4 '14 at 15:17
  • $\begingroup$ I figured it would be easier in cylindrical coordinates, but I'm still curious if what I've done so far (although it's more difficult) is correct? $\endgroup$ – Joshua Jun 4 '14 at 15:28
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@Joshua: Clumsy as your solution is :), it's almost right. There won't be four pieces, just two. You're integrating over a trapezoid in $uv$-space for each piece.

First, we have to decide how $\Sigma$ is oriented. Is the normal pointing outwards? Your cross-product should have been $\left(\dfrac{-u}{\sqrt{1-u^2}},-1,0\right)$, which points inward rather than outward. You need to change the sign (formally, switch $u$ and $v$ to get the right orientation).

Your integral should be [note your $v$ limits were wrong and there's a sign problem on the cross-product] $$\int_{-1}^1 \int_0^{u+2} \left( \frac{2u^2}{\sqrt{1-u^2}} - 3\sqrt{1-u^2}\right) \,dv\,du.$$ Then switching the sign on $y$ will parametrize the other part. Again, be careful about the normal. Add the two.

By the way, your answer should be $-2\pi$. :)

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  • $\begingroup$ So when adding two integrations I just need to make sure my normal should point in the "right" way? (i.e. either both outwards or both inwards?) Thanks for your help! $\endgroup$ – Joshua Jun 5 '14 at 12:31
  • $\begingroup$ Yes, you need the orientation everywhere consistent with what is specified. $\endgroup$ – Ted Shifrin Jun 5 '14 at 12:33

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