# Centroid of a triangle using Green's theorem

1. Apr 12, 2009

### compliant

1. The problem statement, all variables and given/known data
Given a curve C that starts from the origin, goes to (1,0) then goes to (0,1), then back to the origin, find the centroid of the enclosed area D.

2. Relevant equations
$$\bar{x} = {1/(2A)}*\int_C {x^2 dy}$$

$$\bar{y} = -{1/(2A)}*\int_C {y^2 dx}$$

3. The attempt at a solution
Well, obviously the path of C is that of a triangle, and the area is 1/2, which means I have, for x-bar and y-bar, $${1/4}\int_C {x^2 dy}$$ and $${1/4}\int_C {y^2 dx}$$ respectively. My problem is: what are the upper and lower bounds for the line integral?

Last edited: Apr 12, 2009
2. Apr 12, 2009

### Dick

I don't think you can do it as a line integral. Why don't you set it up as a area integral dx*dy? That's the usual way.

3. Apr 12, 2009

### compliant

Because I'm only supposed to use those equations to solve the problem.

4. Apr 12, 2009

### Billy Bob

Do you mean evaluate the line integral directly? Break C up into three segments.

C1 goes (0,0) to (1,0).

C2 goes (1,0) to (0,1).

C3 goes (0,1) to (0,0).

Calculate each of the three line integrals separately, then add.

Do you want to parametrize each one separately? Example: C3 is x=0, y=1-t, for 0<t<1.

5. Apr 12, 2009

### Dick

Then you have to break the integral up into three parts. Integrate over each side of the triangle separately. And your integrands are wrong. You integrate x^2 and y^2 over the area dA=dx*dy to get the moment. If you are going to do it as a contour you need different integrands.

Last edited: Apr 13, 2009