What is the area of $[r = \frac{4}{2 - \cos \theta}]$?

Question

It makes an ellipse, but I'm unsure where to go from here.

Answer 1

To express it in rectangular form, the objective is to rewrite your equation in terms of $x$ and $y$ variables.

Rearrange algebraically your equation to obtain $$2r - r \cos \theta=4$$ Then use the facts of $x=r \cos \theta$ and $r^2=x^2+y^2$.

Answer 2

The area can be computed in polar coordinates as $$ \int_0^{2\pi}\frac12r^2\,\mathrm{d}\theta=\int_0^{2\pi}\frac12\left(\frac4{2-\cos(\theta)}\right)^2\,\mathrm{d}\theta $$ If you prefer rectangular coordinates $$ x=r\cos(\theta)=\frac{4\cos(\theta)}{2-\cos(\theta)} $$ and $$ y=r\sin(\theta)=\frac{4\sin(\theta)}{2-\cos(\theta)} $$ Then $$ 1 =\frac{3x^2}{16}-\frac x2+\frac{y^2}4 =\frac3{16}\left(x-\frac43\right)^2-\frac13+\frac{y^2}4 $$ So the ellipse is $$ \frac9{64}\left(x-\frac43\right)^2+\frac{3y^2}{16}=1 $$ One can use the formula $\pi ab$ for the area, where $a$ and $b$ are the semi-major and semi-minor axes.