Surface integral

  • Thread starter squenshl
  • Start date
  • #1
squenshl
479
4
Let S be a parametrised surface given by (x, y, z) = R(u, v) := (u2, v2, u + v), for 0 [tex]\leq[/tex] u [tex]\leq[/tex] 1 and
0 [tex]\leq[/tex] v [tex]\leq[/tex] 1. How do I find the integral K := [tex]\int\int_S[/tex] z/2 dA.
 

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
Gold Member
4,790
20
By definition, this integral is

[tex]\int_0^1\int_0^1 \frac{u+v}{2}\sqrt{E(u,v)G(u,v)-(F(u,v))^2}dudv[/tex]

where
[tex]E=R_u\cdot R_u[/tex]
[tex]F=R_u\cdot R_v[/tex]
[tex]G=R_v\cdot R_v[/tex]

And this you know how to do.
 
  • #3
squenshl
479
4
That's easy. Cheers.
 

Suggested for: Surface integral

Replies
0
Views
153
  • Last Post
Replies
5
Views
481
Replies
33
Views
1K
Replies
4
Views
657
  • Last Post
Replies
2
Views
927
Replies
2
Views
711
  • Last Post
Replies
2
Views
341
Replies
1
Views
178
  • Last Post
Replies
3
Views
192
Top