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Math Help - Find centroid of the following double integral

  1. #1
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    Find centroid of the following double integral

    we have SS_D f(x,y) dxdy=




    I have to first change the order of Integration and then find the centroid (x0,y0) of the region D (the center is the center of mass considering constant mass density).

    Maybe its an easy question but i'm just starting to study the subject so I didn't get very far in my attempt to solve this.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Find centroid of the following double integral

    Quote Originally Posted by damule View Post
    we have SS_D f(x,y) dxdy=

    I have to first change the order of Integration and then find the centroid (x0,y0) of the region D (the center is the center of mass considering constant mass density).
    The mass of the region

    D \equiv\begin{Bmatrix} 0\leq y \leq 3\\\dfrac{y}{\sqrt{3}}\leq x\leq \sqrt{12-y^2}\end{matrix}

    with density f(x,y) is

    M=\iint_{D}f(x,y)\;dxdy=\int_{0}^3\left(\int_{y/\sqrt{e}}^{\sqrt{12-y^2}}f(x,y)\;dx\right)dy

    and the centroid is

    (x_0,y_0)=\dfrac{1}{M}\left(\iint_{D}xf(x,y)\;dxdy  ,\iint_{D}yf(x,y)\;dxdy\right)

    Why do you want to change the order of integration? Have you had any computation difficulties?
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  3. #3
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    Re: Find centroid of the following double integral

    I have the following solution which I didn't understand completely (the bold part):

    The region is the sector of the circle of radius √(12) centered at the origin between the lines y = 0 and x = y/√(3). You can change the order of integration, but it's a really bad idea from a practical stand point. The order dydx requires two integrals because the upper curve changes. You'd have the limits

    0 ≤ x ≤ √(3) and 0 ≤ y ≤ √(3)x for the first integral and

    √(3) ≤ x ≤ √(12) and 0 ≤ y ≤ √(12 - x) for the second integral.

    The mass of the region is the area of a sector of a circle of radius √(12) and inscribed angle π/3. This is

    m = (√(12)) (π/3) = 2π.


    The circle and the line y = 3 meet at the point (√(3), 3) and so the inscribed angle Θ satisfies

    tanΘ = 3/√(3) = √(3) ==> Θ = π/3.

    The moments about the x and y axes are

    3 √(12-y)
    ∫ ∫ x dx dy = 12 = M_y
    0 y/√3

    3 √(12-y)
    ∫ ∫ y dx dy = 4√(3) = M_x.
    0 y/√3

    The centroid is (M_y/m, M_x/m) = (6/π, 2√(3)/π).



    How did they find that m? On what do you rely? I hope someone could explain to me in further detail

    Is there like 2 ways to calculate the mass? Why didn't he do the double integral to find M?
    Last edited by damule; December 20th 2011 at 12:34 PM.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Find centroid of the following double integral

    Quote Originally Posted by damule View Post
    How did they find that m? On what do you rely? I hope someone could explain to me in further detail
    A=\iint_D\;dxdy is the area of the region D . If f(x,y)=k (constant) the mass of D is m=k\iint_D\;dxdy=kA ( A area of D , which is a sector). You can choose k=1 for finding the centroid ( k appears multiplying in the numerator and denominator and can be canceled).
    Last edited by FernandoRevilla; December 21st 2011 at 02:08 AM.
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