Area bounded by a region in ellipse.

Question

Consider a standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a>b$.

I need to find out the area of the region bounded by the ellipse between the lines $x = a.e$ (i.e. line passing through the focus) and $x = a$.

Here $e$ is the eccentricity of the ellipse.

I tried to integrate $y$ as a function of $x$ with the limits $x = ae$ to a but I'm having difficulty in integrating the function.

Answer 1

Solving for $y$ you get $$y=\frac{b \sqrt{a^2-x^2}}{a}$$ Integrating from $x$ of the focus to $a$ gives an improper integral,

So you can get the area by subtracting a quarter of the ellipse

$$\frac{A}{2}=\frac{\pi a b}{4}-\int_0^{\sqrt{a^2-b^2}} \frac{b \sqrt{a^2-x^2}}{a} \, dx$$ where $\sqrt{a^2-b^2}$ is the positive $x$ of the focus.

I got $$A=\frac{\pi a b}{2}-\frac{b^2 \sqrt{a^2-b^2}}{a}-a b \arctan\left(\frac{\sqrt{a^2-b^2}}{b}\right)$$

Answer 2

Hint:

The required area can be given as:

$$A=\int_{x=a\cdot e}^{a} \int_{y=\frac{-b \sqrt{a^2-x^2}}{a}}^{\frac{b \sqrt{a^2-x^2}}{a}} dy dx=2 \int_{x=a\cdot e}^{a}\frac{b \sqrt{a^2-x^2}}{a} dx. $$

Now, put $x=a \cos \theta$ and find the limits for $\theta$, then solve it.