Question

Centroids o_o

Answer 1

y=2x, y=0,x=1
Find the exact location of the centroid
x bar =
y bar =

Answer 2

@zepdrix :>

Answer 3

Are we talking the centroid of a triangle?

If so, it is located at the point of concurrency of the the medians of the triangle.

Are these these median line equations: y=2x, y=0 , x=1

Answer 4

Yes it's a triangle :)

Answer 5

@Directrix

Answer 6

so I did
\[A=\int\limits_{0}^{1}2xdx=1\] is that part correct?

Answer 7

\[xbar=\frac{ 1 }{ A }\int\limits_{0}^{1}x(2x)dx\]\[=\int\limits_{0}^{1}2x^2=\frac{ 2x^3 }{ 3 }|^1_0=\frac{ 2 }{ 3 }\]

Answer 8

\[ybar=\frac{ 1 }{ A }\int\limits_{0}^{1}\frac{ 1 }{ 2 }(2x)^2\]\[=\frac{ 1 }{ 2 }(\frac{ 4x^3 }{ 3 })|^1_0=\frac{ 2 }{ 3 }\]

Answer 9

\[(xbar,ybar)=(\frac{ 2 }{ 3 },\frac{ 2 }{ 3 })\] is that correct?

Answer 10

I do not see a common intersection for these 3 equations:

y=2x, y=0,x=1

I think they may be the equations for the sides of the triangle.

The centroid does touch each median at a 2:1 ratio.

Answer 11

hmm do you recognize those formulas I used?

Answer 12

\(\overline{x} \) looks good to me
y may need little check

Answer 13

hmhm what part of y?

Answer 14

i guess when you simplified

Answer 15

A=2 right?

Answer 16

i mean 1/A =2

Answer 17

(1/A) int (1/2)(2x)^2
pull the 1/2 out in front
\[\frac{ 1 }{ 2 }\int\limits_{0}^{1}(2x)^2dx =\frac{ 1 }{ 2 }\int\limits_{0}^{1}4x^2dx\]\[\frac{ 1 }{ 2 }(\frac{ 4x^3 }{ 3 })\]then evaluate
nono A=1

Answer 18

A= Area under the curve

Answer 19

Right
the integral of 2x evaluated from 0 to 1
= 1

Answer 20

oh right! i was looking at your drawing you did x=1 y=1
confused me it looked like area is 1/2

Answer 21

whoops sorry heh

Answer 22

then you have everything is good :)

Answer 23

Thanks for the help!! :D

Answer 24

no problem