Area of an Ellipse
Area of an Ellipse
Using the definition of area, we can calculate the sum of unit squares for any object with a closed space. In the case of an ellipse, an ellipse does not have a defined radius, but it has a defined semi-major and semi-minor axis.
In the simulation below, you can input the semi-major and semi-minor radii of your choice. Note that there are small unit squares on the graph, which can be counted manually to get the area of the ellipse. Although we would get the correct answer, this is not practical as an ellipse with a large enough semi-major and semi-minor radii would take a long time to count up all the individual unit squares each of length and width 1m by 1m.
Since an ellipse is a two dimensional shape and to plot the semi-major and semi-minor radii require two dimensions, we can assume that an ellipse will be in two dimensions. We can use the equation of a circle x2 + y2 = 1 and include the semi-major and semi-minor radii as a and b. The equation for an ellipse now becomes, x2⁄a2 + y2⁄b2 = 1. From this equation we can solve for y, we can then integrate with respect to the semi-minor radius a. Since the xy plane is divided into 4 quadrants, we need to multiply the area of an ellipse in one quadrant by 4 in order to get the total area of the ellipse.
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
\[ \frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} \]
\[ a^2 \left(\frac{y^2}{b^2}\right) = a^2 \left(1 - \frac{x^2}{a^2}\right) \]
\[ \frac{a^2 y^2}{b^2} = a^2 - \frac{a^2 x^2}{a^2} \]
\[ \frac{a^2 y^2}{b^2} = a^2 - x^2 \]
\[ b^2 \left(\frac{a^2 y^2}{b^2}\right) = b^2 (a^2 - x^2) \]
\[ a^2 y^2 = b^2 (a^2 - x^2) \]
\[ \sqrt{a^2 y^2} = \sqrt{b^2 (a^2 - x^2)} \]
\[ a y = b \sqrt{a^2 - x^2} \]
\[ y = \frac{b}{a} \sqrt{a^2 - x^2} \]
Before we integrate, since the xy plane is divided into 4 quadrants, we need to multiply the area of an ellipse in one quadrant by 4 in order to get the total area of the ellipse.
So now our integral set up should look like this:
\[ A = \int_{0}^{a} 4 y dx \]
\[ A = \int_{0}^{a} 4 \left(\frac{b}{a} \sqrt{a^2 - x^2}\right) dx \]
\[ A = 4 \cdot \frac{b}{a} \int_{0}^{a} \sqrt{a^2 - x^2} dx \]
We can substitute x as a sin(t), and dx as a cos(t) dt.
This will also change the integral bounds to from 0 to a to 0 to π⁄2.
\[ A = 4 \cdot \frac{b}{a} \int_{0}^{\frac{\pi}{2}} \sqrt{a^2 - (a \sin(t))^2} a \cos(t) dt \]
\[ A = 4 \cdot \frac{b}{a} \int_{0}^{\frac{\pi}{2}} \sqrt{a^2 - a^2 \sin^2(t)} a \cos(t) dt \]
\[ A = 4 \cdot \frac{b}{a} \int_{0}^{\frac{\pi}{2}} \sqrt{a^2 (1 - \sin^2(t))} a \cos(t) dt \]
\[ A = 4 \cdot \frac{b}{a} \int_{0}^{\frac{\pi}{2}} a \sqrt{1 - \sin^2(t)} a \cos(t) dt \]
\[ A = 4 \cdot \frac{b}{a} \int_{0}^{\frac{\pi}{2}} a^2 \sqrt{1 - \sin^2(t)} \cos(t) dt \]
\[ A = 4 b a \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \sin^2(t)} \cos(t) dt \]
\[ A = 4 b a \int_{0}^{\frac{\pi}{2}} \sqrt{\sin^2(t) + \cos^2(t) - \sin^2(t)} \cos(t) dt \]
\[ A = 4 b a \int_{0}^{\frac{\pi}{2}} \sqrt{\cos^2(t)} \cos(t) dt \]
\[ A = 4 b a \int_{0}^{\frac{\pi}{2}} \cos(t) \cos(t) dt \]
\[ A = 4 b a \int_{0}^{\frac{\pi}{2}} \cos^2(t) dt \]
\[ A = 4 b a \left( \frac{t}{2} + \frac{\sin(2x)}{4} \right) _{0}^{\frac{\pi}{2}} \]
\[ A = 4 b a \left( \frac{\pi}{4} \right) \]
\[ A = \pi a b \]
Therefore the equation we get to calculate the area of an ellipse is πab.
\[ A = \pi \times a \times b \]
An ellipse has two radii a semi-major and semi-minor which is an enclosed space that is plotted on the xy plane in Cartesian coordinates, or (r-minor, r-major) plane in polar coordinates, so that would mean that the units of an ellipse's area would be m2.
\[ m^2 \]
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