find the center of masses of the plane area bounded by the parabolas y=2x-x^2 and y=3x^2-6X

- anonymous

find the center of masses of the plane area bounded by the parabolas y=2x-x^2 and y=3x^2-6X

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- anonymous

what math class is this from?

- anonymous

the method I see uses calculus

- anonymous

@mbugwav you can't leave me to do this alone

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## More answers

- anonymous

PETER 14 I AGREEit can be solved using calculus

- anonymous

dan815 yes it will what about x bar?

- dan815

i will restart

- dan815

|dw:1378206924244:dw|

- anonymous

https://www.desmos.com/calculator/qcshlnluve
have a graph

- dan815

or maybe 1/ b-a ?

- anonymous

this problem can be solved by using the knowledge the area

- dan815

|dw:1378207580735:dw|

- anonymous

I am going to type the equation just now, then we can take it from there

- anonymous

\[\frac{ \int\limits_{A}^{}xdA }{ \int\limits_{A}^{} dA}\]

- anonymous

peter i appreciate
dan815 what are the centers

- dan815

|dw:1378207611672:dw|

- dan815

does -2/3 for y bar make sense?

- anonymous

what about x bar

- dan815

im thinking that is just 1

- dan815

but what about ybar?? i am trying some sktechy logic

- dan815

lets wait and see how pneew does it

- anonymous

alright

- anonymous

dan how did you get x=1?

- dan815

just avg, they intersection at 0 and 2 so i said 1

- anonymous

dA stand for the derivative of an Area

- anonymous

\[dA=(2x-x^2)-(3x^2-6x)dx\]

- anonymous

sorry guys, I think I have a solution here, maybe if you may consider what i'm trying to post

- anonymous

area =16/3

- anonymous

\[ \frac{ \int\limits_{0}^{2}x(2x-x^2)-(3x^2-6x)dx}{ \int\limits_{0}^{2} ((2x-x^2)-(3x^2-6x)dx}\]

- anonymous

i am getting y bar as -0.4

- anonymous

simple put dA in the equation

- anonymous

then you integrate that, since this is a calculus question

- dan815

alright mbugwav ill use that formula

- dan815

that shud be 3/16

- dan815

let me check your area just in case

- anonymous

yes, sir that shud be 3/16, that will give 1, which means x=1

- dan815

ok

- anonymous

true bcause its the reciprocal but why not\[-4x^2\]

- anonymous

for y, |dw:1378209580611:dw|

- anonymous

now,\[y=(2x-x^2)+(3x^2-6x)\]

- anonymous

i agree

- anonymous

dA is the same, then you put what you have in the equation of y

- dan815

|dw:1378210124670:dw|

- dan815

spotted a mistake

- dan815

anyway go on

- anonymous

|dw:1378210072047:dw|, than you integrate

- dan815

|dw:1378210239272:dw|

- dan815

use this formula

- dan815

http://www.youtube.com/watch?v=H5RcfMIZ_yw

- dan815

use that formula there

- dan815

for y bar

- anonymous

okay

- anonymous

sorry guys i have to go to do my praticals, dan is correct, but i think |dw:1378210681566:dw|, if you can integrate this correctly you find the answer.

- anonymous

thakns guys i have to go

- goformit100

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