Solving Equation of Spheroid for Prolate & Oblate

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SUMMARY

The discussion centers on the equations of prolate and oblate spheroids, specifically the equations derived from rotating an ellipse around different axes. A prolate spheroid is defined by the equation X^2/a^2 + [Y^2+Z^2]/b^2 = 1, where 'a' is the semi-major axis along the X-axis. Conversely, an oblate spheroid has the same equation but with 'b' as the semi-major axis, indicating that both equations appear identical under certain conditions. The participants clarify that to distinguish between the two shapes, one must rotate the ellipse around different axes, leading to different configurations and equations.

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Homework Statement



I have a doubt on spheroid equations. A prolate spheroid is obtained by rotating the ellipse,
X^2/a^2 + Y^2/b^2 = 1 {Here a is major axis}
about the semi-major axis a(i.e. X axis). Its equation is,
X^2/a^2 + [Y^2+Z^2]/b^2 = 1
An oblate spheroid is obtained by rotating the ellipse,
X^2/a^2 + Y^2/b^2 = 1 {Here b is major axis}
about the semi-minor axis a(i.e. X axis).Its equation is,
X^2/a^2 + [Y^2+Z^2]/b^2 = 1
The problem is that both equations are identical. What I have done is that I have taken ‘a’ always along X axis and ‘b’ always along Y axis. Is it necessary that the equations be distinguishable?

Homework Equations



X^2/a^2 + Y^2/b^2 = 1

The Attempt at a Solution



In order distinguish between the two, I will have to take ‘a’ along Y axis for one of them. Suppose I take ‘a’ along the Y axis for oblate spheroid case, the equation of the oblate spheroid is got by rotating the ellipse,
X^2/b^2 + Y^2/a^2 = 1
about the semi-minor axis ‘b’(i.e. X axis).Its equation is,
X^2/b^2 + [Y^2+Z^2]/a^2 = 1
Another way is to rotate the ellipse,
X^2/a^2 + Y^2/b^2 = 1 {Here a is major axis}
first along X axis(i.e. ‘a’) for prolate spheroid in which case the equation becomes,
X^2/a^2 + [Y^2+Z^2]/b^2 = 1
And then along Y axis(i.e. ‘b’) for oblate spheroid in which case the equation becomes,
[X^2+Z^2]/a^2 + Y^2/b^2 = 1 {Here b is major axis}

Is there a better way to this? Please help.
 
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Start with the general ellipsoid:
\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2 = 1

if it is rotationally symmetric around anyone principle axis the the (a,b,c) coefficients for the other two must be equal. Example: if you have rotational symmetry about x then b=c. Further in this example if a is much bigger than b=c then the ellipsoid is cigar shaped. If a is much smaller than b=c then it is "cow-pie" shaped.

Regards,
J. Baugh
 
jambaugh said:
Further in this example if a is much bigger than b=c then the ellipsoid is cigar shaped. If a is much smaller than b=c then it is "cow-pie" shaped.

Could you please say what are the shapes of Cigar and Cow-pie in this context?
 
Amith2006 said:
Could you please say what are the shapes of Cigar and Cow-pie in this context?

Sure, take a highly eccentric ellipse. Rotate about the long axis and you have an ellipsoid that is long and cylindrical like a tapered cigar.

Take the same ellipse and rotate about the short axis and you have a tapered disk shaped ellipsoid, like the shape of a discus used in track-n-field events or like the pile of defecant a cow leaves behind.

Regards,
James Baugh
 
So, the equation of prolate spheroid is,
X^2/a^2 + [Y^2+Z^2]/b^2 = 1
for a>b=c
and the equation of oblate spheroid is also,
X^2/a^2 + [Y^2+Z^2]/b^2 = 1
but here a<b=c. I that what u meant?
 

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