From polar coordinates to Cartesian.

Question

Please help me move from polar coordinates to Cartesian. $p=\sqrt{\cos(\pi + \phi})$. I tried to do that and made following steps, but I don't know if I'm right.

Answer 1

Let's limit $\theta$ to $[0,2\pi)$. Therefore$$r=\sqrt{-\cos\theta}\qquad,\qquad \dfrac{\pi}{2}\le\theta\le\dfrac{3\pi}{2}$$and we have$$\sqrt{(x^2+y^2)^3}=-x$$

Answer 2

First try polar plot to grasp what is involved. The sign of quantity under radical sign is positive in quadrants 2 and 3.

Square root includes either sign even if not explicitly given. So all quadrants should be occupied. We have to square the equation to include both signs.

$$ r^2=-\cos \theta = \frac{-x}{r}$$

$$ r^3= -x$$

To include both signs square both sides. $$ r^6= (x^2+y^2)^3 =x^2$$

So this is the cartesian form without change if $-x$ is plugged in place of $x$ or $-y$ is plugged in place of $y.$

There is symmetry with respect to both axes forming a 2 leaved bean shape.