Learn how to find the area of an ellipse using polar coordinates with detailed step-by-step derivations and trigonometric substitution.
Equation of Ellipse in Polar Coordinates
The equation of an ellipse centered at the origin is given by:
\[ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \qquad (I) \]The conversion from rectangular to polar coordinates is given by:
\[ x = r \cos \theta, \quad y = r \sin \theta \]Substitute the conversion formulas into equation (I):
\[ \dfrac{r^2 \cos^2 \theta}{a^2} + \dfrac{r^2 \sin^2 \theta}{b^2} = 1 \]Divide all terms in the above equation by \( \cos^2 \theta \) and simplify:
\[ \dfrac{r^2}{a^2} + \dfrac{r^2 \tan^2 \theta}{b^2} = \sec^2 \theta \] \[ r^2 \left( \dfrac{1}{a^2} + \dfrac{\tan^2 \theta}{b^2} \right) = \sec^2 \theta \] \[ r^2 \left( \dfrac{b^2 + a^2 \tan^2 \theta}{a^2 b^2} \right) = \sec^2 \theta \]Solve for \( r^2 \):
Area of Ellipse in Polar Coordinates
The area \( A \) in polar coordinates enclosed by a curve is given by the formula:
\[ A = \dfrac{1}{2} \int_0^{2\pi} r(\theta)^2 \, d\theta \]Substitute \( r^2 \) of the ellipse found above into the area integral:
\[ A = \dfrac{1}{2} \int_0^{2\pi} a^2 b^2 \dfrac{\sec^2 \theta}{b^2 + a^2 \tan^2 \theta} \, d\theta \qquad (II) \]Use the trigonometric substitution method that transforms \( a \tan \theta \) into \( b \tan \alpha \) so that the denominator can be factored and further simplifications occur:
\[ b \tan \alpha = a \tan \theta \qquad (III) \]Differentiate both sides with respect to \( \theta \), using the chain rule on the left side:
\[ b \sec^2 \alpha \dfrac{d\alpha}{d\theta} = a \sec^2 \theta \]Which gives:
\[ \sec^2 \theta \, d\theta = \dfrac{b}{a} \sec^2 \alpha \, d\alpha \]Or:
\[ d\theta = \dfrac{\dfrac{b}{a} \sec^2 \alpha}{\sec^2 \theta} \, d\alpha \qquad (IV) \]Limits of integration:
From (III) above, we may write \( \alpha = \arctan \left( \dfrac{a}{b} \tan \theta \right) \).
- For \( \theta = 0 \), \( \alpha = 0 \)
- For \( \theta = 2\pi \), \( \alpha = 2\pi \)
Substitute \( d\theta \) and the limits of integration into the integral:
\[ A = \dfrac{1}{2} \int_0^{2\pi} a^2 b^2 \dfrac{\sec^2 \theta}{b^2 + b^2 \tan^2 \alpha} \left( \dfrac{\dfrac{b}{a} \sec^2 \alpha}{\sec^2 \theta} \right) \, d\alpha \]Factor \( b^2 \) in the denominator, take all constants outside the integral, and simplify:
\[ A = \dfrac{ab}{2} \int_0^{2\pi} \dfrac{1}{1 + \tan^2 \alpha} \sec^2 \alpha \, d\alpha \]Use the trigonometric identity \( 1 + \tan^2 \alpha = \sec^2 \alpha \) in the denominator and simplify:
\[ A = \dfrac{ab}{2} \int_0^{2\pi} d\alpha \]Evaluate the integral:
\[ A = \dfrac{ab}{2} \left[ \alpha \right]_0^{2\pi} \]Simplify to obtain the final formula for the area of an ellipse: