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Thread: help starting a triple integral/center of mass problem

  1. #1
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    Question help starting a triple integral/center of mass problem

    i need help setting up the integral expressions to compute the centroid please!
    the solid bounded by the paraboloids z=1+x^2+y^2 and z=5-x^2-y^2 with density proportional to the distance from the plane z=5
    so i need to find the boundaries for x,y, and z, which i think i can do if i set x=0 then y=0?
    my main problem is figuring out how to put the density into an equation. i think it has to do with the distance formula. f(x,y,z)= k* sqrt of 5? any hints would help a ton! thanks!
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  2. #2
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    I tried to do this problem last night but ended up getting swamped by calculations. One way to find the answer is to switch to cylindrical coordinates:

    $\displaystyle \begin{aligned}
    x&=r\cos\theta\\
    y&=r\sin\theta\\
    z&=z
    \end{aligned}$

    so that the equations for the paraboloids become

    $\displaystyle z=1+r^2$
    $\displaystyle z=5-r^2.$

    Solving for the intersection of the two gives us

    $\displaystyle \begin{aligned}
    1+r^2&=5-r^2\\
    2r^2&=4\\
    r^2&=2\\
    r&=\sqrt{2}
    \end{aligned}$

    for the radius of a circle in the $\displaystyle xy$-plane that bounds the solid. Our center of mass for the $\displaystyle z$-coordinate becomes

    $\displaystyle z_M=\frac{1}{V}\int_0^{2\pi}\int_0^{\sqrt{2}}\int_ {1+r^2}^{5-r^2} z\cdot k(5-z)\cdot r\,dz\,dr\,d\theta,$

    where $\displaystyle V$ is the volume of the solid.

    Hint: The $\displaystyle x$- and $\displaystyle y$-coordinates of the center of mass must be zero. Why is this?
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