# Area of Ellipse in Polar Coordinates

   

## 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 $A = a b \pi$