# Thread: Solving a problem without the use of stokes

1. ## Solving a problem without the use of stokes

Hey all!

My professor in my vector analysis course likes to always have a two part question where one part can be solved by stokes, greens or gauss theorem. They are pretty straight forward if you understand them and I find it that I have. The other part is where I take the same problem and solve it without the use of the theorem.

So here is an example I've solved for stokes, but I don't know how to solve without it.

The Question
We have a triangle with vertris (1,0,0) (0,1,0) (0,0,1).
solve the closed integral of [int] xy dx+yz dy+zx dz over that plane.

Attempt at solution
We know the plane equation is x+y+z=1
We also know [int] xy dx+yz dy+zx dz = [int] F dot dr, which is a closed line integral.

This is pretty much as far as I get..

What I need help with
I have a vague memory that I can devide C into parts of integration, but as I have just one integral and can only integrate over x,y or z seperatly.

Please give me a few hints so I can carry on with my understanding of calculus

2. ## Re: Solving a problem without the use of stokes

For example, if $\displaystyle A=(1,0,0),\;B=(0,1,0)$ then, $\displaystyle AB\equiv (x,y,z)=(1-t,t,0)$ with $\displaystyle t\in [0,1]$ so

$\displaystyle \int_{AB}xydx+yzdy+zxdz=\int_0^1(1-t)t(-dt)=\ldots$

Idem for the other sides.

3. ## Re: Solving a problem without the use of stokes

Originally Posted by FernandoRevilla
For example, if $\displaystyle A=(1,0,0),\;B=(0,1,0)$ then, $\displaystyle AB\equiv (x,y,z)=(1-t,t,0)$ with $\displaystyle t\in [0,1]$ so

$\displaystyle \int_{AB}xydx+yzdy+zxdz=\int_0^1(1-t)t(-dt)=\ldots$

Idem for the other sides.
Aha, so parametization is the way to go. So I do it beween AB, AC and CB.

Thank you very much. Im going to post alot here these days because I really need help with this

4. ## Re: Solving a problem without the use of stokes

I might suggest that you can use the plane x + y + z = 1. On each of your three lines, one of x, y or z = 0. I would also sugest that you set up all three integral before evaluating. Two of the three may cancel leaving only one, or, because of the actual plane in this example, you could combine the three integrals in a single integral.