# Centroid of a triangle using Green's theorem

## Homework Statement

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.

## Homework Equations

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

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

## 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?

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Dick
Homework Helper
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.

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

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.

Dick