Centroid of the region bounded by the curve

[johnq2k7]

centroid of the region bounded by the curve...need help!

Find the centroid of the region bounded by the curve x=2-y^2 and the y-axis:

my work shown:

therefore if A= 2 times the integral of sqrt(2-x) dx

is the M_x equal to the integral of (2-x) dx from 0 to 2?

and the M_y equal to the integral of (2)(x)(sqrt(2-x) dx from 0 to 2?

therefore x-coordinate of the centroid is M_y/A

and the y-coordinate of the centroid is M_x/A

therefore centroid is [(M/y/A),(M_x/A)]

is this correct?

then the x-coordinate of the centroid is (M_y / A)

i've been told the centroid of the y-coord. is zero... .however i dont' believe that is correct.. how do i determine the centroid and are M_x and M_y values correct... because if they are ... isn't the centroid simply x--> M_y/A and y--> M_x/A... please help me with this problem!

 

[Dr.D]

Why not work out your integrals on y instead of x, -2<=y<=+2?

Then just work the problem out and see what you get without worrying in advance what it is supposed to be.

 

[johnq2k7]

Why not work out your integrals on y instead of x, -2<=y<=+2?

Then just work the problem out and see what you get without worrying in advance what it is supposed to be.

i'm confused with what u mean.. do u mean integrate the equation in terms of x instead of y to determine the centroid.. how do i go about doing that...

isn't my A(y) value correct, i just need help with my values for M_y, M_x assuming my A(y) is correct therefore i could find the centroid as M_y/A,M_x/A for co-ord. of (x,y)

please help

 

[Dr.D]

The given boundary function is x = f(y), so you can calculate the area and the moments as integrals that involve things like
A = int(x) dy from -2 to 2, etc.