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]

 

 

As we know,

 

[tex]\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[/tex]

 

where

[tex]\nabla \times F(x,y,z) = P i + Q j + R k[/tex] and [tex] g = f(x,y) = z[/tex]

 

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, [tex]\frac{\partial g}{\partial x} = 1, \frac{\partial g}{\partial y} = 0[/tex]

 

So correcting my arithmetic..

 

[tex]F \cdot \nabla f = -(1 - x - y) + y + z - 2y dA[/tex]

= [tex]\int_0^1 \int_0^1 -1 + x + z [/tex]

 

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

 

[tex]\int_0^1 \int_0^1 2x - 1 dx dy = \int_0^1 x^2 - x (1,0) = \int_0^1 0 ....[/tex]

 

what am I doing wrong here?

 

argh, does this forum support latex?

Your LaTeX looks pretty good, this forum uses the [ math ] and [ / math ] tags (without spaces) to display LaTeX. change all those [tex] tags and I think that your post would be a lot easier to read -- and then I think I can help you with Stokes' theorem, too.

 

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

 

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

 

[math]F \cdot dr = (1-x-y) i + (1-2z) j + (y+z-2y) k.[/math]

 

 

As we know,

 

[math]\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[/math]

 

where

[math]\nabla \times F(x,y,z) = P i + Q j + R k[/math] and [math] g = f(x,y) = z[/math]

 

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, [math]\frac{\partial g}{\partial x} = 1, \frac{\partial g}{\partial y} = 0[/math]

 

So correcting my arithmetic..

 

[math]F \cdot \nabla f = -(1 - x - y) + y + z - 2y dA[/math]

= [math]\int_0^1 \int_0^1 -1 + x + z [/math]

 

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

 

[math]\int_0^1 \int_0^1 2x - 1 dx dy = \int_0^1 x^2 - x (1,0) = \int_0^1 0 ....[/math]

 

what am I doing wrong here?

 

argh, does this forum support latex?

 

let's see how that looks... probably more like you intended.

 

Firstly, I don't think that you can ignore the y dependence, since the points go from (0,0,0) to (1,1,1).

 

Secondly, I think that this is going to involve multiple line integrals, since you are going to want to go each point separately. I.e. the total line integral equals the line integral from (0,0,0) to (1,0,1) plus the line integral from (1,0,1) to (1,1,1) plus the other two. The nice thing about breaking it into each part is that you can calculate the normal to each line pretty easily.