ellipsoid

**putnam120** · November 21st 2006, 04:47 PM

Find with proof the volume of the region that is contained within $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ and with $a,b,c \neq 0$ .

Sure I know this can be done with calculus (multiple integrals or volume from rotation) but if that were the kind of solution I was looking for I would have posted in the Calculus section.

Hint: Orthogonal Projections

**galactus** · November 21st 2006, 05:04 PM

Here's one method:

Each perpendicular slice to the z-axis for |z|<c is an ellipse whose equation is

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

or

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

The area of which is:

${\pi}\left(\frac{a}{c}\sqrt{c^{2}-z^{2}}\right)\left(\frac{b}{c}\sqrt{c^{2}-z^{2}}\right)$ $={\pi}\frac{ab}{c^{2}}\left(c^{2}-z^{2}\right)$ so $V=2\int_{0}^{c}{\pi}\frac{ab}{c^{2}}(c^{2}-z^{2})dz=\frac{4}{3}{\pi}abc$

**putnam120** · November 21st 2006, 05:14 PM

nice solution.
does anyone else have one. preferably one that does not use integrals.

**ThePerfectHacker** · November 21st 2006, 05:32 PM

Originally Posted by putnam120

nice solution.
does anyone else have one. preferably one that does not use integrals.

I do not think there is another way let me explain. Usually these argument use "almost close" arguements. That means you subdivide your region into simpler rectangular regions and you add them. If you do use a different way you need to prove convergence (a fact that physics and engineering classes avoid). For example, you cannot say the increasing the number of sides in a regular polygon inscribed in a circle get closer to the area of a circle, you need to justify that. The way you do that is by using the error bounds. You need to show that the error bound converge to zero and just assume. Using integration avoids this problem, if your function is continous then the Riemann sum always exists and is the area below the curve (or the volume). So integration is the preferred way.

If you want I can show you how these approximation arguments lead to a problem.

I have a different problem. What is the perimeter of an ellipse?

**putnam120** · November 21st 2006, 06:39 PM

ok here is how it is done without calculus.

before i start i must first find the area of the largest triangle that will fit into an ellipse with major axis of length 2a and minor axis of length 2b also note that the formula for the area of an ellipse is $ab\pi$ where 2a and 2b are the length of the axies. using orthogonal projectiles we make the ellipse into a circle with radius of length b. since this kind of projection preserves ratios of areas we have

$\frac{(ABC)}{b^2\pi}=\frac{(xyz)}{ab\pi}$ (ABC) is the area of the largest triangle that can fit into the circle, this area is $\frac{3b^2\sqrt{3}}{4}$ (i assume that you can prove this on your own). (xyz) is the area of the larges triangle that can fit into the ellipse. so solving for (xyz) we have

$(xyz)=\frac{3ab\sqrt{3}}{16}$ .

now for the original problem.
WLOG assume that $a\le b\le c$

once again we make a projection that converts the ellipsoid into a sphere of radius a. now consider the area of the largest tetrahedron that fits into the upper half of the sphere. the area of this is (ABCD). $(ABCD)=\frac{a^3\sqrt{3}}{4}$

now think about (wxyz) the area of the larges tetrahedron that will fit into the upper half of the ellipse, with base on the face where c=0. it is plain to see that the volume of this is expressed as $(wxyz)=\frac{abc\sqrt{3}}{16}$

now using the fact the ratio of volumes are preserved we have

$\frac{\frac{a^3\sqrt{3}}{4}}{\frac{4}{3}a^3\pi}=\f rac{\frac{abc\sqrt{3}}{16}}{V}$

solving for $V$ we have

$V=\frac{4}{3}abc\pi$

and V is the volume of the ellipsoid.

like i said in the hint, use orthogonal projections, and calculus was not needed.

**ThePerfectHacker** · November 21st 2006, 06:45 PM

Originally Posted by putnam120

now using the fact the ratio of volumes are preserved we have

I did not really read what you are doing. But when I saw that my trained eye caught an error. I assume you are talking about the ratio's of volumes is the same as the cube of the sides. But how do you know that is true? A fact people learn but is never proven.

**putnam120** · November 21st 2006, 07:20 PM

its proven in almost any geometry book that covers, distortions and projections.

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