Volume of an Ellipsoid
Volume of an Ellipsoid
Using the definition of volume, we can calculate the sum of unit squares for any object with a closed space. In the case of an ellipsoid, an ellipsoid has a defined radius.
In the simulation below, you can input an arbitrary radius of your choice. To calculate the volume you can count each individual unit cubes to get the volume of an ellipsoid. Although we would get the correct answer, this is not practical as an ellipsoid with a large enough radius would take a long time to count up all the individual unit cubes each of length, width and height 1m by 1m by 1m.
Since an ellipsoid is a three dimensional representation of an ellipse, and to plot the three radii require three dimensions, we can assume that an ellipsoid will be in three dimensions. Since every ellipsoid has three radii a, b, c, we can take the integral of the radii to find the volume of an ellipsoid.
We can first start out by writing the equation of an ellipsoid:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \]
Solving for the ellipsoid to be in the yx plane we get:
\[ \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 - \frac{x^2}{a^2} \]
Dividing by 1 - x2⁄a2, we get:
\[ \frac{y^2}{b^2 \left( 1 - \frac{x^2}{a^2} \right)} + \frac{z^2}{c^2 \left( 1 - \frac{x^2}{a^2} \right)} = 1 \]
The semi-axes become:
\[ b_1 = b \left( \sqrt{1 - \frac{x^2}{a^2}} \right) \text{ and } c_1 = c \left( \sqrt{1 - \frac{x^2}{a^2}} \right) \]
The area of an ellipse is πabc.
In this case it is written as πb1c1 . Which can be solved the following way using the b1 and c1 equations above:
\[ A = \pi b_1 c_1 \]
\[ A = \pi \left( b \left( \sqrt{1 - \frac{x^2}{a^2}} \right) \right) \left( c \left( \sqrt{1 - \frac{x^2}{a^2}} \right) \right) \]
\[ A = \pi b c \left( 1 - \frac{x^2}{a^2} \right) \]
We can now take the integral of the area of an ellipse, which we can then calculate the integral and find the volume of an ellipsoid:
\[ A = \pi abc \]
\[ V = \int_{-a}^{a} \pi abc dx \]
\[ V = \int_{-a}^{a} \pi b c \left( 1 - \frac{x^2}{a^2} \right) dx \]
\[ V = \pi b c \int_{-a}^{a} \left( 1 - \frac{x^2}{a^2} \right) dx \]
\[ V = \pi b c \left( x - \frac{x^3}{3a^2}_{-a}^{a} \right) \]
\[ V = \pi b c \left( (a - -a) - \left( \frac{a^3}{3a^2} - -\frac{a^3}{3a^2} \right) \right) \]
\[ V = \pi b c \left( 2a - \frac{2a^3}{3a^2} \right) \]
\[ V = \pi b c \left( \frac{6a}{3} - \frac{2a}{3} \right) \]
\[ V = \pi b c \left( \frac{4a}{3} \right) \]
\[ V = \frac{4}{3} \pi a b c \]
An equation we can create therefore to calculate the volume of an ellipsoid are the three radii multiplied together times four thirds times π.
\[ V = \frac{4}{3} \pi a b c \]
Since an ellipsoid is an enclosed space in three dimensions, that would mean that the units would be meters cubed.
\[ m^3 \]
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