centroid of a hemisphere

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June 4th 2009, 10:39 PM
manyarrows

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
June 4th 2009, 11:58 PM
matheagle

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 $0\le \phi \le \pi/2$ .
June 5th 2009, 12:20 AM
Random Variable

$dV = \pi r^{2}dz$ , where r is the radius of each circular disc. Then you to write r in terms of z.
June 5th 2009, 12:29 AM
manyarrows

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
June 5th 2009, 12:47 AM
manyarrows

I found my mistake dV should be

$dV=\pi y^2 dz$

then

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

thanks again
June 5th 2009, 06:35 AM
matheagle

I read dV as dxdydz. Changing to spherical with $J=\rho^2\sin\phi$ and $z=\rho\cos\phi$ we have

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

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

$= { (a^4/4)(2\pi)(1/2) \over (a^3/3)(2\pi)(1)}={3a\over 8}$ .
June 5th 2009, 07:45 AM
manyarrows

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
June 5th 2009, 08:44 PM
matheagle

Quote:

Originally Posted by manyarrows

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.
June 5th 2009, 09:24 PM
Random Variable

He did it correctly the second time. You don't have to set up a triple integral.
April 19th 2012, 01:01 PM
DrewMeadow

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