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Identity

Member

Registered: 2007-04-18

Posts: 934

surface integral with stokes

, C is the positively oriented boundary as viewed from above of the part of the plane

in the first octant.

Use Stokes' Theorem to evaluate

.

Stokes' Theorem:

Can someone please show me the steps you would take, since I'm still quite confused about surface integration
thanks

Last edited by Identity (2010-01-07 21:28:07)

rzaidan

Member

Registered: 2009-08-13

Posts: 59

Re: surface integral with stokes

Hi Identity :
you are to find curl F which is the crosss product of del × F to get
del × F = 3x i -(3y-x) j +2y k
then find the normal to the surface to get n = (1/√11)(3i+j+k)
then find ds = dydx/(n.k)=√11 dx dy
afterthat apply stokes  theorem:
∫F.dr = ∫ ∫del  × F .n ds =1/√11 ∫ ∫ 10 x -y)dydx   on R where R is the tringle x + 2y = 3 that is the integral limits are : x from 0 to 3-2y and y from 0 to 3/2
and complete
I think it will be easy
Best Wishes
Riad Zaidan

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: surface integral with stokes

Thankyou rzaidan, tha helped