How to find the area of this polar graph that seems to be an ellipse.

Question

You have the polar graph $\dfrac{4}{2-\cos \theta}$ that seems to be an ellipse. I just don't really know how to find the area of the ellipse of this equation.

Answer 1

The ellipse with focal origin can be expressed as:

$$r=\frac{b^2}{a \pm c\cos \theta} \quad \text{ or } \quad r=\frac{b^2}{a \pm c\sin \theta}$$

where $a^2=b^2+c^2$

Now $$\frac{a}{b^2}=\frac{2}{4} \quad \text{ and } \quad \frac{c}{b^2}=\frac{1}{4}$$

We have $$\frac{(a+c)(a-c)}{b^4}=\frac{3}{4} \times \frac{1}{4} \implies b=\frac{4}{\sqrt{3}} \quad \text{and} \quad a=\frac{8}{3}$$

The area is $$\pi a b=\dfrac{32\pi}{3\sqrt{3}}$$

Answer 2

So, we basically have $r(\theta) = \frac{4}{2-\cos\theta}$ as the equation of the curve.

Now, area under curve is given by $$A = \int_0^{2\pi} \frac{1}{2}r^2d\theta$$ $$\implies A = \int_0^{2\pi}\frac{8}{(2-\cos\theta)^2}d\theta$$

Now, this integral can be solved by making using the tan half-angle substitution for cos, and then another substitution, and we get the following

$$A = \frac{32\pi}{3\sqrt3}$$