# centroid of a hemisphere

• Jun 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.

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

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

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

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

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

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

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

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

Thanks
• Jun 4th 2009, 11:58 PM
matheagle
These are supposed to be triple integrals.
You should switch to spherical co-ordinates with the region being
$\displaystyle 0\le \rho\le a$, $\displaystyle 0\le \theta \le 2\pi$ and $\displaystyle 0\le \phi \le \pi/2$.
• Jun 5th 2009, 12:20 AM
Random Variable
$\displaystyle dV = \pi r^{2}dz$, where r is the radius of each circular disc. Then you to write r in terms of z.
• Jun 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

$\displaystyle a^2=y^2 + z^2$

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

Thanks again
• Jun 5th 2009, 12:47 AM
manyarrows
I found my mistake dV should be

$\displaystyle dV=\pi y^2 dz$

then

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

thanks again
• Jun 5th 2009, 06:35 AM
matheagle
I read dV as dxdydz. Changing to spherical with $\displaystyle J=\rho^2\sin\phi$ and $\displaystyle z=\rho\cos\phi$ we have

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

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

$\displaystyle = { (a^4/4)(2\pi)(1/2) \over (a^3/3)(2\pi)(1)}={3a\over 8}$.
• Jun 5th 2009, 07:45 AM
manyarrows
I also did it this way

$\displaystyle hemisphere = a^2=y^2+x^2$

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

$\displaystyle dV= \pi r^2$

$\displaystyle r=y$

$\displaystyle \tilde{z}=z$

$\displaystyle y^2=a^2-z^2$

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

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

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

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

Your way looks much easier, so thanks
• Jun 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

$\displaystyle a^2=y^2 + z^2$

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

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

The other bounds of integration would be $\displaystyle -\sqrt{a^2-x^2}\le y \le \sqrt{a^2-x^2}$ and $\displaystyle -a\le x \le a$. Which is begging for polar, i.e., trig substitution.
• Jun 5th 2009, 09:24 PM
Random Variable
He did it correctly the second time. You don't have to set up a triple integral.
• Apr 19th 2012, 01:01 PM