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

  1. #1
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    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)

    I'm sure I may have made some mistakes in my integration set up for A,M_x,M_y please help

  2. #2
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    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
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  3. #3
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    Re: centroid of the region bounded by the curve..work show..help

    Quote 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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