Question

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

Answer 1

`+-(x^2+y^2)^(3/2)=4y`

Explanation:

Given that

`x^2+y^2=r^2`

`y=rsin\theta`

`sin\theta=y/r`

`r=sqrt(x^2+y^2)`

We proceed as follows

`x^2+y^2=4(y/r)`

`x^2+y^2=(4y)/sqrt(x^2+y^2)`

`(x^2+y^2)(x^2+y^2)^(1/2)=4y`

`(x^2+y^2)^(2/2)(x^2+y^2)^(1/2)=4y`

`(x^2+y^2)^(3/2)=4y`

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

`-(x^2+y^2)^(3/2)=4y`

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

`(r^2)^(3/2)=4rsin\theta`

`r^3=4rsin\theta`

Dividing both sides by `r`

`r^2=4sin\theta`