# Surface integrals

1. Feb 8, 2007

### bodensee9

[Can someone help with the following? I am supposed to evaluate the line integral of ∫F∙dr. The curve is oriented counterclockwise as viewed from above. So suppose that F(x,y,z) = (x+y^2)I + (y+z^2)j + (z+x^2)k, and C is the triangle formed by (1,00), (0,1,0), (0,0,1).

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.

2. Feb 8, 2007

### mathwonk

i cant read all that, but couldn't you just check your work by simply parametrizing the triangle and just doing a one variable integral?

it is especially easy on the vertical and horizontal dies where, on the y axis side say, both dx and dz are zero.

3. Feb 8, 2007

### bodensee9

But I'm supposed to evaluate it using the double integral, so.

4. Feb 8, 2007

### arildno

The plane equation is, indeed, x+y+z=1.
Hence, the normal vector is parallell to (1,1,1)

5. Feb 8, 2007

### bodensee9

Oh I see where I went wrong! Thanks!!!