Ask Me Help Desk - Centroid of an ellipsoid

What is the centroid (or center of mass) of a homogenous half-ellipsoid in terms of its semi-axes (a, b, c)? Is it different from that of a half-ellipse?

The center of mass is at the intersection of its semi-major and semi-minor axes.

But where in the z axis? Remember this is half-ellipsoid.

Quote:

Originally Posted by jimmy1843

Remember this is half-ellipsoid.

Sorry - I missed that. You have values for a, b, c, for the equation

$ \frac {x^2} {a^2} + \frac {y^2}{b^2} + \frac {z^2}{c^2} = 1 $

Right? The center of mass along the z-axis can be found from

$ z_{cm} = \frac {\int_0 ^c zA(z) dz}{V} $

where V = volume of the half elipsoid, which is $V= \frac 2 3 \pi abc$ , and A(z) is the cross-sectional area as a function of z. The cross-section is an ellipse - you will need to come up with equations for the lengths of the semi-major an semi-minor axes as a function of z, use that to find an expression for A as a function of z (hint - the area of an ellipse is $\pi$ times the lengths of the semi-major and semi-minor axes), and put that into the above integral. It works out pretty nicely - post back with what you get.

Quote:

Originally Posted by ebaines

Sorry - I missed that. You have values for a, b, c, for the equation

$ \frac {x^2} {a^2} + \frac {y^2}{b^2} + \frac {z^2}{c^2} = 1 $

Right? The center of mass along the z-axis can be found from

$ z_{cm} = \frac {\int_0 ^c zA(z) dz}{V} $

where V = volume of the half elipsoid, which is $V= \frac 2 3 \pi abc$ , and A(z) is the cross-sectional area as a function of z. The cross-section is an ellipse - you will need to come up with equations for the lengths of the semi-major an semi-minor axes as a function of z, use that to find an expression for A as a function of z (hint - the area of an ellipse is $\pi$ times the lengths of the semi-major and semi-minor axes), and put that into the above integral. It works out pretty nicely - post back with what you get.

In order to findan expression for A as a function of z, we consider:
pi*x*y=A(z)

We use:
$ \frac {x^2} {a^2} + \frac {z^2}{c^2} = 1 $

and
$ \frac {y^2}{b^2} + \frac {z^2}{c^2} = 1 $

solving for x and y in terms of z, we obtain:
x = a*sqrt(1 - z^2/c^2)
y = b*sqrt(1 - z^2/c^2)

putting these two in the integral, we come up with
$ z_{cm} = \frac {\int_0 ^c pi*x*y dz}{V} $

Substituting x and y:
$ z_{cm} = \frac {\int_0 ^c pi* a*sqrt(1 - z^2/c^2)*b*sqrt(1 - z^2/c^2) dz}{V} $

After simplifying and integrating we get:

$ z_{cm} = -3/4*piabc^2/{2/3*pi*abc} $

$ z_{cm} = -9/8c $

Am I on the right track?

Quote:

Originally Posted by jimmy1843

x = a*sqrt(1 - z^2/c^2)
y = b*sqrt(1 - z^2/c^2)

putting these two in the integral, we come up with
$ z_{cm} = \frac {\int_0 ^c pi*x*y dz}{V} $

Not quite - the formula is:

$ z_{cm} = \frac { \int_0 ^c zdV} {V} = \frac {\int_0 ^c z \pi xy dz}{V} $

Note the 'z' term that you left out. It results in the numerator becoming

$ \int_0 ^c \pi ab z(1-\frac {z^2}{ c^2}) dz $

Quote:

Originally Posted by jimmy1843

$ z_{cm} = -9/8c $

Am I on the right track?

Clearly the value for $z_{cm}$ must lie somewhere between 0 and z. With the correction above you should get it now.

Quote:

Originally Posted by ebaines

Not quite - the formula is:

$ z_{cm} = \frac { \int_0 ^c zdV} {V} = \frac {\int_0 ^c z \pi xy dz}{V} $

Note the 'z' term that you left out. It results in the numerator becoming

$ \int_0 ^c \pi ab z(1-\frac {z^2}{ c^2}) dz $

Clearly the value for $z_{cm}$ must lie somewhere between 0 and z. With the corection above you should get it now.

Oh yes! Sorry I missed it. My oversight. I think the answer should be:
$z_{cm}$ = 3/8c
I hope this time I'm correct (?)
Thanks.

Quote:

Originally Posted by jimmy1843

I think the answer should be:
$z_{cm}$ = 3/8c
I hope this time I'm correct (?)
Thanks.

Yes - I agree with your answer - nicely done!