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.

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.

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?

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







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.

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.

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!