Where have I gone wrong with this surface integral problem?

[Lucy Yeats]

Homework Statement

Evaluate the surface integral ∫F.dS where F = xi - yj + zk and where the surface S is of the cylinder defined by x^2+y^2≤4, and 0≤z≤1. Verify your answer using the Divergence Theorem.

Homework Equations

The Attempt at a Solution

I parametrized the surface in terms of θ and z: r=(2cosθ, 2sinθ, z). I found dr/dθ X dr/dz=(2cosθ, 2sinθ, z). (How do I know which way round to do the cross product?). I rewrote F as (2cosθ, -2sinθ, z). I then did the following integral:
∫∫(2cosθ, -2sinθ, z).(2cosθ, 2sinθ, z)dθdz and got ∫∫4(cosθ)^2-4(sinθ)^2 dθ dz=0

But using the divergence theorem: divF=1, so I found the answer to be the volume of the cylinder, 4pi.

Where have I gone wrong?

Thanks in advance! :-)

 

[Office_Shredder]

Did you include the top and bottom surfaces of the cylinder?

 

[Lucy Yeats]

Ah, I forgot those... Thanks for pointing that out. I'll try again now and see if I get the right answer. Could someone explain how you work out which way round to do the cross products?

 

[Lucy Yeats]

Thanks, I got the right answer. :-)

I'm still confused about the cross product thing I mentioned before though...

 

[HallsofIvy]

Your original problem, "Evaluate the surface integral ∫F.dS where F = xi - yj + zk and where the surface S is of the cylinder defined by x^2+y^2≤4, and 0≤z≤1" isn't complete. You need to specify the orientation of the surface as well. Here, as for general closed surfaces, you can specify either "oriented by outward pointing normals" or "oriented by inward pointing normals".

Changing the order of a cross product only changes the direction of the normal vector. In your example, you get <2cos(\theta), 2sin(\theta), z> If you check different values of \theta, for example, for \theta= 0 so that is <2, 0, z>or \theta= \pi/2 so that is <0, 2, z> you can see that the vector is pointing <b>away<b> from the the z-axis, which is the axis of the cylinder, and so is "outward pointing". If you did the cross product in the other order, you would get <-2cos(\theta) -2sin(\theta), -z> so that every vector is pointing <b>toward</b> the axis. That normal is "inward pointing" and integrating with that would give the negative of the integral using the "outward pointing" normal.<br /> <br /> That also, by the way, means that when you integrate over the two ends of the cylinder, you must make the same "outward" or "inward" choice. If you used outward pointing normals for the cylinder, then on the plane z= 1, you must use dS= <0, 0, 1> dxdy so it is pointing upward (out of the cylinder) and on the plane z= 0 you must use dX= <0, 0, -1>dxdy pointing downward (but still out of the cylinder).</b></b>

 

[Lucy Yeats]

Thanks so much; this makes perfect sense now. :-)