Area of Ellipse in Polar Coordinates

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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 rectangualar to polar coordinates is given by \[ x = r \cos \theta \quad \text{and} \quad y = r \sin \theta \] Substitute the above in equation (I) \[ \dfrac{(r \cos \theta)^2}{a^2} + \dfrac{(r \sin \theta)^2}{b^2} = 1 \] Divide all terms in the above equation by \( \cos^2 \theta \) and simplify \[ \dfrac{r^2}{a^2} + \dfrac{r^2 \tan^2 \theta }{b^2} = \sec^2 \theta \] Solve for \( r^2 \) \[ r^2 = a^2 b^2 \dfrac{\sec^2 \theta}{b^2+a^2\tan^2 \theta} \]



Area of Ellipse in Polar Coordinates

The area \( A \) in polar coordinate enclosed by a curve is given by the formula \[ A = \dfrac{1}{2} \int_0^{2\pi} r(\theta)^2 d \theta \] where \( r(\theta) \) is the equation of the curve in polar coordinates,
Substitute \( r^2 \) of the ellipse found above \[ A = \dfrac{1}{2} \int_0^{2\pi} a^2 b^2 \dfrac{\sec^2 \theta}{b^2+a^2\tan^2 \theta} d\theta \qquad (II) \] Use the
trigonometric substitution method that transform \( \quad a \tan \theta \) into \( b \tan \alpha \quad \) so that the denominator can be factored and hence more simplifications occur.
Let \[ b \tan \alpha = a \tan \theta \qquad (III) \] Differentiate both sides of the above with respect to \( \theta \), using the
chain rule of differentiation 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 \( \alpha = \arctan (\dfrac{a}{b} \tan \theta ) \)
For \( \theta = 0 \) , \( \alpha =0 \)
For \( \theta = 2\pi \) , \( \alpha = 2\pi \)
Substitute \( d\theta \) in (IV) and the limits of integration in the integral \[ A = \dfrac{1}{2} \int_0^{2\pi} a^2 b^2 \dfrac{\sec^2 \theta}{b^2+b^2\tan^2 \alpha} \dfrac{\dfrac{b}{a} \sec^2 \alpha}{\sec^2 \theta } \; d \alpha \] Factor \( b^2 \) in the denominator and take all constant outside the integral and simplify \[ A = \dfrac{ a b}{2} \int_0^{2\pi} \dfrac{1}{1+\tan^2 \alpha} \sec^2 \alpha \; d \alpha \] Use the
trigonometric identity \( 1+1\tan^2 \alpha = \sec^2 \alpha \) in the denominator and simplify \[ A = \dfrac{ a b}{2} \int_0^{2\pi} d \alpha \] Evaluate the integral \[ A = \dfrac{ a b}{2} \left[\alpha\right]_0^{2\pi} d \alpha \] Simplify \[ A = a b \pi \]



More References and Links

  1. Equation of an Ellipse
  2. Area in Polar Coordinates
  3. Polar Coordinates and Equations
  4. Polar Coordinates.

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