- #1

- 178

- 0

I know that the integral ∫F∙dr over some surface S is equal to ∫∫curlF∙dS and is equal to ∫∫curlF∙n/|n|ds or ∫∫curlF∙ndA, where n is the normal vector. So, curl F is -2zi -2xj -2yk. And since the region has the expression x+y+z = 1, so suppose the region z can be written as (x, y, 1-x-y), and the normal vector to this region has the expression <dz/dx, -dz/dy, 1> or <-1,-1,1>. So wouldn’t curl F dot n be 2z+2x-2y, and since z = 1-x-y, curl F n should be 2-2x-2y+2x-2y? So I am supposed to integrate the integral ∫∫2-4y dydx where 0≤y≤1-x and 0≤x≤1? Can someone point out what I’m doing wrong here? Thanks.