Nevermind, I got it.
c = Mc/M, where Mc is the double integral over the region R [c(dA)].
Since the intersections are at (0, 0), (4, 0), and (2, 2), it's easiest to set this up in dydx.
For x this becomes integral(0, 2) integral (y^2/2, 4-y) x dydx. Divide this by the integral over the same region dydx. The process for y is identical, except the for My, integrate over the function y instead of x.
(x, y) = (64/35, 5/7)
Find the centroid of the region in the first quadrant bounded by the x-axis, the parabola y^2 = 2x, and the line x + y = 4.
I've graphed the function, and it looks like a triangle with one side curved (the parabola). I'm not quite sure how to go about the rest of the problem though.
2 answers
A quick calculation shows that the line intersects the parabola at (2,2) in the first quadrant
lots of work ahead of us:
first you must find the area of the region, it has to be done in two parts:
from x=0 to x=2 under the curve, then
from x=2 to x=4 under the straight line
The rest is best illustrated by watching the youtube video
https://www.youtube.com/watch?v=p0XUlLPpL2Q
Make sure you watch parts 1 and 2
lots of work ahead of us:
first you must find the area of the region, it has to be done in two parts:
from x=0 to x=2 under the curve, then
from x=2 to x=4 under the straight line
The rest is best illustrated by watching the youtube video
https://www.youtube.com/watch?v=p0XUlLPpL2Q
Make sure you watch parts 1 and 2
1 answer