Stokes Theorem

  • Thread starter HeheZz
  • Start date
  • #1
11
0

Homework Statement



Verify Stokes Theorem ∬(∇xF).N dA where surface S is the paraboloid z = 0.5(x^2 + y^2) bound by the plane z=2, Cis its boundary, and the vector field F = 3yi - xzj + yzk.


The Attempt at a Solution



I had found (∇xF) = (z+x)i + (-z-3)k

r = [u, v, 0.5(u^2 + v^2)]
Therefore N= ru X rv = -ui -uj +k
Therefore (∇xF).N = [(z+x), 0, (-z-3)].[-x, -y, 1]
After that I substitute x = rcos(θ), y = rsin(θ), z = 0.5r^2
Thus ∫(0-2)∫(0-2pi) (∇xF).Nr dθdr

But I cant seems to get the answer. Can anyone help? Help would greatly appreciated :)
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,228
1,830
Looks okay. What did you get for the answer or where are you getting stuck?
 
  • #3
11
0
ok so its ∫(0-2)∫(0-2pi) (∇xF).Nr dθdr

Therefore its:
∫(0-2)∫(0-2pi) -xz-x^2-z-3 dA (Using x = rcos(θ), y = rsin(θ), z = 0.5r^2)
= ∫(0-2)∫[(0-2pi) -0.5r^3cosθ - r^2cos^2(θ) - 0.5r^2 - 3]r dθdr

Im not sure is my steps correct? Jus a little problem with integrating cos^2(θ).

Furthermore, I've used ∫F.dr to do a check and the answer seems to be 0. Is this correct?
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,228
1,830
Looks good, but I got -20 pi for the answer both ways.

To integrate cos^2(θ), use the trig identity cos^2(θ)=[1+cos(2θ)]/2.
 

Related Threads on Stokes Theorem

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
16
Views
5K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
625
Top