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Math Help - centroid of a hemisphere

  1. #1
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    centroid of a hemisphere

    I must find the z centroid of a hemisphere with radius a. It's base is on the x-y plane and its dome extends up the z axis. I am using the following equations to determine the centroid.

    \overline{z}=\frac{\int_V\tilde{z} dV}{\int_V dV}

    I am using dV=\pi a^2 dz and \tilde{z}=z and integrating from 0 to a

    \overline{z}=\frac{\int_{0}^{a} z \pi a^2 dz}{\int_{0}^{a} \pi a^2 dz}

    \overline{z}=\frac{\pi a^2 \int_{0}^{a} z dz}{\pi a^2\int_{0}^{a} dz}

    \overline{z}=\frac{ \int_{0}^{a} z dz}{\int_{0}^{a} dz}

    \overline{z}=\frac{\frac{z^2}{2}|_{0}^{a}}{z|_{0}^  {a}}

    \overline{z}=\frac{a}{2}

    I know the answer is supposed to be \overline{z}=\frac{3a}{8} Where did I mess up?

    Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    These are supposed to be triple integrals.
    You should switch to spherical co-ordinates with the region being
    0\le \rho\le a, 0\le \theta \le 2\pi and 0\le \phi \le \pi/2.
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  3. #3
    Super Member Random Variable's Avatar
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     dV = \pi r^{2}dz , where r is the radius of each circular disc. Then you to write r in terms of z.
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  4. #4
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    The hemisphere is center on the origin so I know that the x and y centroid are 0. Is this what you are refering to by triple integrals?

    I also would like to be able to do this in cartesian if that is possible as that was that coordinates the problem specified.

    If I write a in terms of z I get

    a^2=y^2 + z^2

    Then when ever I try to get the integrand in terms of one variable I wind back up at a^2

    Thanks again
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  5. #5
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    I found my mistake dV should be

    dV=\pi y^2 dz

    then

    dV= \pi (a^2 -z^2) dz

    thanks again
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  6. #6
    MHF Contributor matheagle's Avatar
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    I read dV as dxdydz. Changing to spherical with J=\rho^2\sin\phi and z=\rho\cos\phi we have

     { \int_0^a\int_0^{2\pi}\int_0^{\pi /2} \rho^3\sin\phi \cos\phi d\phi d\theta d\rho\over  \int_0^a\int_0^{2\pi}\int_0^{\pi /2} \rho^2 \sin\phi d\phi d\theta d\rho}

    = { \int_0^a \rho^3d\rho \int_0^{2\pi}d\theta \int_0^{\pi /2} \sin\phi \cos\phi d\phi \over  \int_0^a \rho^2 d\rho \int_0^{2\pi}d\theta \int_0^{\pi /2} \sin\phi d\phi }

    = { (a^4/4)(2\pi)(1/2) \over  (a^3/3)(2\pi)(1)}={3a\over 8}.
    Last edited by matheagle; June 5th 2009 at 08:39 PM.
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  7. #7
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    I also did it this way

    hemisphere = a^2=y^2+x^2

    \tilde{z}=\frac{\int_{V}\tilde{z} dV}{\int_{V} dV}

    dV= \pi r^2

    r=y

    \tilde{z}=z

    y^2=a^2-z^2

    \tilde{z}=\frac{\int_{0}^{a}z \pi y^2 dz}{\int_{0}^{a}\pi y^2 dz}

    \tilde{z}=\frac{\int_{0}^{a}z \pi (a^2-z^2) dz}{\int_{0}^{a}\pi (a^2-z^2) dz}

    \tilde{z}=\frac{\pi \int_{0}^{a}z (a^2-z^2) dz}{\pi \int_{0}^{a}(a^2-z^2) dz}

    \tilde{z}=\frac{\int_{0}^{a}z (a^2-z^2) dz}{ \int_{0}^{a}(a^2-z^2) dz}

    Your way looks much easier, so thanks
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by manyarrows View Post
    The hemisphere is center on the origin so I know that the x and y centroid are 0. Is this what you are refering to by triple integrals?

    I also would like to be able to do this in cartesian if that is possible as that was that coordinates the problem specified.

    If I write a in terms of z I get

    a^2=y^2 + z^2

    Then when ever I try to get the integrand in terms of one variable I wind back up at a^2

    Thanks again
    You can do this via (x,y,z) but to solve the integrals you will need to various trig substitutions. It's smarter to switch to spherical immediately.

    For example the bound for z would be 0\le z\le \sqrt{a^2-x^2-y^2}. Then the (x,y) base is a circle of radius a, also screaming out for trig substitution.

    The other bounds of integration would be -\sqrt{a^2-x^2}\le y \le \sqrt{a^2-x^2} and -a\le x \le a. Which is begging for polar, i.e., trig substitution.
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  9. #9
    Super Member Random Variable's Avatar
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    He did it correctly the second time. You don't have to set up a triple integral.
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  10. #10
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    Re: centroid of a hemisphere

    Here's the problem worked in a bit more detail, hope it reads:

    Calculus: Centroid of a Hemisphere, Math 251
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