How to convert $r^2 = sintheta$ from polar to rectangular form?

Question

Through a bit of questionable math, I got $x^6+x^4y^2+y^4x^2+y^6-x^2$ , but that's definitely not right, and I'm not sure what to do now. Any and all help is greatly appreciated.

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Answer 1 · 2018-05-13T21:44:11

I'm not sure what I can possibly do with my equation either

$r^2= sintheta$

Multiply both sides by $r$ :
$r^2*r= r*sintheta$

Since $rsintheta= y$ and $r^2= x^2+y^2$ :

$(x^2+y^2)*r= y$

And $r= +-sqrt(x^2+y^2)$

First let's move the $r$ over to the left side by dividing by $r$ on both sides:

$(x^2+y^2)= y/r$

$(x^2+y^2)= y/(+-sqrt(x^2+y^2))$

Square both sides to see where this goes:

$(x^2+y^2)^2= (y/(+-sqrt(x^2+y^2)))^2$

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

$y^2=(x^4+2x^2y^2+y^4)(x^2+y^2)$

$y^2= x^6+x^4y^2+2x^4y^2+2y^4x^2+y^4x^2+y^6$

Gave me this graph
graph{y^2= x^6+x^4y^2+2x^4y^2+2y^4x^2+y^4x^2+y^6 [-10, 10, -5, 5]}