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

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Answer 1 · 2015-10-21T17:09:41

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

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 $r$

$r^2=4sin\theta$

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