Thread: centroid of the region bounded by the curve..work show..help

1. centroid of the region bounded by the curve..work show..help

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

A= integral of (2-y^2)dy from 0 to 1

M_y= (1/2) integral of (2-y^2)^2 dy from 0 to 1

M_x= integral of (y)(2-y^2)dy from 0 to 1

x= (M_y)/A
y= (M_x)/A

centroid is (x,y)

2. Re: centroid of the region bounded by the curve..work show..help

What you have is a parabola. The limits are 0 to 2.

Here is a diagram of the region you have to find the centroid of.

By symmetry, y=0 is the y coordinate of the centroid.

See here: http://mathworld.wolfram.com/ParabolicSegment.html

3. Re: centroid of the region bounded by the curve..work show..help

Originally Posted by galactus
What you have is a parabola. The limits are 0 to 2.

Here is a diagram of the region you have to find the centroid of.

By symmetry, y=0 is the y coordinate of the centroid.

See here: http://mathworld.wolfram.com/ParabolicSegment.html
Therefore is the following approach correct?:

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

A= integral of (sqrt(2-x)dx from 0 to 2

M_y= (1/2) integral of (sqrt(2-x))^2 dx from 0 to 2

M_x= integral of (x)(sqrt(2-x))dx from 0 to 2

x= (M_y)/A
y= (M_x)/A

centroid is (x,y)

I believe my values for A,M_x,M_y, and the centroid (x,y) should be correct now.. did I make any mistakes in my assumptions or integral setup?

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