- #1
Simfish
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Homework Statement
Let C be the closed curve that goes along straight lines from (0,0,0) to (1,0,1) to (1,1,1) to (0,1,0) and back to (0,0,0). Let F be the vector field [tex]F = (x^2 + y^2 + z^2) i + (xy + xz + yz) j + (x + y + z)k. Find \int F \cdot dr[/tex]
By Stokes Theorem, I know that I can transform this into a curl (instructions say don't evaluate the line integral directly).
So I know I get a form of square that goes CCW. Projected on the xz-plane, it is simply a square (but the y-coordinate of the right end is 1).
I think the plane is described by x = z, since there is no dependence on y. So x-z = 0
Anyways, I take the curl of F to get
[tex]F \cdot dr = (1-x-y) i + (1-2z) j + (y+z-2y) k.[/tex]
I take the dot product of this with the plane z - x = 0 to get
[tex] \nabla f = -i + k [/tex]
and dot that with F to get
[tex]F \cdot \nabla f = x + y -1 + y + z - 2y = x + z -1 [/tex]
so then
[tex]\int_0^1 \int_0^1 (x + z -1) dx dy.[/tex] Since x = z...
[tex]\int_0^1 \int_0^1 dx dy = \int_0^1 x^2 - x dy = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6}[/tex]
am I doing this correctly?