# Math Help - Stokes' theorem

1. ## Stokes' theorem

I got stuck using the Stokes' theorem, the problem is at the bottom of the pic. I found the Curl of F, and also the normal of the Triangle. As you can see, I ended up with an area integer with 3 variables, how do I solve this? did I do it right? http://img2.tapuz.co.il/forums/1_119833460.jpg

2. Originally Posted by asi123
I got stuck using the Stokes' theorem, the problem is at the bottom of the pic. I found the Curl of F, and also the normal of the Triangle. As you can see, I ended up with an area integer with 3 variables, how do I solve this? did I do it right? http://img2.tapuz.co.il/forums/1_119833460.jpg
A couple of things:

1. Since you're integrating on the surface (a plane), you can substitute z = 4x + y (which is the equation of the plane). This solves your technical infelicity of three variables in the integrand ......

2. It looks like you're using a unit vector normal to the surface (plane). That's incorrect. $F \cdot dS \neq F \cdot \hat{n} \, dx \, dy$ in general.

In fact, $F \cdot dS = F \cdot n \, dx \, dy$:

$F \cdot dS = F \cdot \left( \frac{\partial z}{\partial x} \, i + \frac{\partial z}{\partial y} \, j - k\right) \, dx \, dy$.

3. Are you OK with the region in the xy-plane you integrate over (and hence the integral terminals) .....? The region is the right angle triangle with vertices at (0, 0), (2, 0) and (2, 2).

3. Originally Posted by mr fantastic
A couple of things:

1. Since you're integrating on the surface (a plane), you can substitute z = 4x + y (which is the equation of the plane). This solves your technical infelicity of three variables in the integrand ......

2. It looks like you're using a unit vector normal to the surface (plane). That's incorrect. $F \cdot dS \neq F \cdot \hat{n} \, dx \, dy$ in general.

In fact, $F \cdot dS = F \cdot n \, dx \, dy$:

$F \cdot dS = F \cdot \left( \frac{\partial z}{\partial x} \, i + \frac{\partial z}{\partial y} \, j - k\right) \, dx \, dy$.

3. Are you OK with the region in the xy-plane you integrate over (and hence the integral terminals) .....? The region is the right angle triangle with vertices at (0, 0), (2, 0) and (2, 2).
Ok, I think I fixed it, is that right?

Another thing.
I marked everything in the pic.
Shouldn't I normalize the normal vector?
Did I write the boundaries right?

10x again.

4. Originally Posted by asi123
Ok, I think I fixed it, is that right?

Another thing.
I marked everything in the pic.
Shouldn't I normalize the normal vector? Mr F says: No. I've already given the correct formula.
Did I write the boundaries right? Mr F says: I'm not sure why you wanted to switch from xy to uv-coordinates ..... x and y are fine. Using xy-coordinates, the boundary is y = 0 to y = -x + 2 and x = 0 to x = 2 (or alternatively, x = 0 to x = 2 - y and y = 0 to y = 2).

10x again.
I get 20/3 (and I stayed with xy-coordinates) but I could be wrong since my careless arithmetic errors are legion.

5. Originally Posted by mr fantastic
I get 20/3 (and I stayed with xy-coordinates) but I could be wrong since my careless arithmetic errors are legion.
Why did you choose the boundaries from y=0 to y=-x+2, isn't it from y=0 to y=x?

6. Originally Posted by asi123
Why did you choose the boundaries from y=0 to y=-x+2, isn't it from y=0 to y=x?
Quite right. I accidently read my (2, 2) as (0, 2) when I drew my sketch ......