Volume of ellipsoid

[simplethinker]

I've been trying to derive the formula V=4/3*πabc for an ellipsoid, but I can't get it right. Here's what I've done:
The basic equation of an ellipsoid is

I used the substitutions
and got which is the same as

And this is what I did:

Somewhere I lost an "abc" term, but I'm not sure where.

I also tried with another form of an ellipsoid
and the substitutions to get
By using the same techniques as before, I ended up with
and, remembering the substitutions, I have
which has two extra "abc" terms.

What am I doing wrong? I've read ahead a few chapters (okay, maybe a dozen), so my teacher isn't of much help. I've gone in-depth into a lot of techniques and the rationale behind them, I understand what I'm doing by finding the tripe-iterated integral, so it's not just me following the steps outlined in the book, but I don't see what I'm doing wrong.

[DarkerLine]

You're not following the substitution rule correctly. When you substitute u=x/a, you forget to substitute du = dx/a which will give you an extra factor of a at the beginning; similarly the substitutions v=y/b and w=z/c will give you extra factors of b and c.

In the second example, then, you'll have factors of 1/ab, 1/ac, and 1/bc which will cancel out two of the (abc) terms.

A much more interesting miscalculation is in the surface area of a sphere (or an ellipsoid - you'll do the same substitutions anyway). If you decided to do it using rectangular coordinates, you'll get the following integral: [attachment=1954:attachment]

This integral evaluates to Pi^2 instead of 4 Pi (the correct answer for a sphere of radius 1). Why?

[thornahawk]

simplethinker:

I'm glad that the Equation Editor was useful (if you did use it to generate those expressions), so allow me to add a tip the next time you use LaTeX for generating math expressions: "\arcsin(x)" renders better than "arcsin(x)" (without the backslash) in LaTeX. Barring that, you could always enclose function names in "\mathrm{}", like "\mathrm{arcsin}(x)". :)

DarkerLine:

I don't follow where you got that integral in the derivation of the sphere volume; that would of course return π² since you're multiplying 2π with the area of a radius-1 semicircle.

thornahawk

[DarkerLine]

That was an integral for the surface area (of a sphere of radius 1 centered at 0) not the volume. x goes from -1 to 1, and what's inside the integral is the circumference of the slice of the circle (parallel to the yz plane) that lies on the sphere's surface and is centered at (x,0,0). The radius of that circle is $\sqrt{1-x^2}$, and the circumference is $2\pi r$

Intuitively, these circumferences should "add up" to be the sphere's surface area.

...would a picture help?

(one of my friends showed this to me a couple of years ago as part of his argument that calculus is fundamentally flawed. I don't think I agree with the argument, but this example is interesting)

[simplethinker]

[DarkerLine]

I meant the radius of the circle, not of the sphere.

(if you think you see a better solution, set up a different integral for the surface area - but without polar coordinates, as that would defeat the point)


Edit: also the integral of y=√(r^2-x^2) with respect to x would give you 1/2 πr^2 not 2πr.

[simplethinker]

I'm not using polar coordinates. The equation of a sphere is r^2=x^2+y^2+z^2, the r I'm using is from that, not polar definitions.
But, here are two ways to find the surface area:


which becomes

and

which becomes

which both evaluate to "4πr^2"

The first is integrating the circumference of the trace of the sphere in the yz-plane from [-r,r] with respect to x.
The second is finding the surface area of the surface of revolution formed by "y=√(r^2-x^2)" being revolved around the y-axis from [-r,r]

[edit] I fixed the error in my last post.

[another edit] Thanks for helping me in my first problem, I knew it was something stupid/simple :biggrin:

[DarkerLine]

Wow. One thing I never realized (although it's nothing esoteric) is that the surface area of the sphere is the same as the lateral surface area of a cylinder circumscribed (is this correct for 3D figures?) about it. This is intuitive and counter-intuitive at the same time.

[simplethinker]

Yeah, that was the way that my math book explained it

@thornahawk: thanks for the tip (I did use LaTeX, it's a nice program )