1. ## stokes theorem

may i know what mistake have i made?

given that the vector field F(x,y,z) = < -y,x,x-2y>. use stokes theorem to compute the upward flux of curl F across z=1-x^2 -y^2 with z>0.

what i did:

curl F =<-2,-1,2>

using stoke theorem, integral F. dr = double integral curl F dS = triple integral grad (Curl F) dxdydz =0 as grad (curl F)=0

may i know why this doesnt work?

thanks!

2. using stoke theorem, integral F. dr = double integral curl F dS
this integral is along the closed curve dr, dS boundary is the closed curve dr.
dS may be arbitrary. (there is no volume)

double integral curl F dS = triple integral grad (Curl F) dxdydz =0
this integral is along volume dxdydz and surfase dS surounds this all volume.

3. does that mean that i cant convert it to triple integral because there is no volume?

sorry im still abit lost about why i cant convert it from double to triple integral. could you explain it to me again?

then when would i know when i can convert from single to double integrals and then from double to triple and when i should not convert ?

thanks so much for your help

4. You can't convert it to a triple integral because Stoke's theorem says nothing about triple integrals! Stoke's theorem says that a certain integral over a surface is equal to another integral around the boundary of that surface.