# How do you convert r^2 = 4 sin θ into a cartesian equation?

Oct 21, 2015

$\pm {\left({x}^{2} + {y}^{2}\right)}^{\frac{3}{2}} = 4 y$

#### Explanation:

Given that

${x}^{2} + {y}^{2} = {r}^{2}$

$y = r \sin \setminus \theta$

$\sin \setminus \theta = \frac{y}{r}$

$r = \sqrt{{x}^{2} + {y}^{2}}$

We proceed as follows

${x}^{2} + {y}^{2} = 4 \left(\frac{y}{r}\right)$

${x}^{2} + {y}^{2} = \frac{4 y}{\sqrt{{x}^{2} + {y}^{2}}}$

$\left({x}^{2} + {y}^{2}\right) {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} = 4 y$

${\left({x}^{2} + {y}^{2}\right)}^{\frac{2}{2}} {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} = 4 y$

${\left({x}^{2} + {y}^{2}\right)}^{\frac{3}{2}} = 4 y$

NOTE: If we want that part of the graph below the x axis it is

$- {\left({x}^{2} + {y}^{2}\right)}^{\frac{3}{2}} = 4 y$

Which corresponds to $r$ going in the other direction in polar coordinates

We can check by converting back and see if we get what we started with

${\left({r}^{2}\right)}^{\frac{3}{2}} = 4 r \sin \setminus \theta$

${r}^{3} = 4 r \sin \setminus \theta$

Dividing both sides by $r$

${r}^{2} = 4 \sin \setminus \theta$