- #1
compliant
- 45
- 0
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
[tex]\bar{x} = {1/(2A)}*\int_C {x^2 dy}[/tex][tex]\bar{y} = -{1/(2A)}*\int_C {y^2 dx}[/tex]
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, [tex]{1/4}\int_C {x^2 dy}[/tex] and [tex]{1/4}\int_C {y^2 dx}[/tex] respectively. My problem is: what are the upper and lower bounds for the line integral?
Last edited: