Stokes Theorem Problem: Evaluating Line Integral with Vector Field

[Simfish]

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 $$F = (x^2 + y^2 + z^2) i + (xy + xz + yz) j + (x + y + z)k. Find \int F \cdot dr$$

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

$$F \cdot dr = (1-x-y) i + (1-2z) j + (y+z-2y) k.$$

I take the dot product of this with the plane z - x = 0 to get

$$\nabla f = -i + k$$

and dot that with F to get
$$F \cdot \nabla f = x + y -1 + y + z - 2y = x + z -1$$

so then

$$\int_0^1 \int_0^1 (x + z -1) dx dy.$$ Since x = z...

$$\int_0^1 \int_0^1 dx dy = \int_0^1 x^2 - x dy = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6}$$

am I doing this correctly?

 

[HallsofIvy]

Part of your LaTex didn't come through. How do you take the dot product of a vector with a plane?

And what do you mean by "dot that with F"?

You have $$F \cdot \nabla f$$

That confuses me greatly. You have a function "F" but say nothing about "f". If you meant $F \cdot \nabla F$, that is not at all what you want.
Stokes theorem has
$$\int\int \nabla \times \vec{F}\cdot \vec{n}dA$$

 

[Simfish]

Okay, here you have the dot product of a vector with the gradient of the plane:
http://tutorial.math.lamar.edu/classes/calcIII/stokestheorem.aspx
(example 2). But this could be confusing, and there is an alternative way to express Stokes Theorem after the dot product has been taking.

This is Stokes Theorem as a surface integral. (the gradient approach to the surface integral is at http://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx )

As we know,

$$\int\int \nabla \times \vec{F}\cdot \vec{n}dA = \int\int F \cdot dS = \int\int -P \frac{\partial g}{\partial x} - Q \frac{\partial g}{\partial y} + R$$

where
$$\nabla \times F(x,y,z) = P i + Q j + R k$$ and $$g = f(x,y) = z$$

The equation of my plane should be z = f(x,y) = x, right? (the plane is f, and determined by "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).")

in that case, $$\frac{\partial g}{\partial x} = 1, \frac{\partial g}{\partial y} = 0$$

So correcting my arithmetic..

$$F \cdot \nabla f = -(1 - x - y) + y + z - 2y dA$$
= $$\int_0^1 \int_0^1 -1 + x + z$$

We take z = x. Is this an appropriate choice? If so

$$\int_0^1 \int_0^1 2x - 1 dx dy = \int_0^1 x^2 - x (1,0) = \int_0^1 0 ...$$

what am I doing wrong here?

 

[simongjh]

You set $$g = f(x,y) = z$$, so your D will be the projection of graph g at x-y plane.
You can substitute $$z=x$$ at there. I got zero as well for this question.