How to convert $r^{2} = sin θ$ $r^2 = sintheta$ from polar to rectangular form?

Question

Through a bit of questionable math, I got $x^{6} + x^{4} y^{2} + y^{4} x^{2} + y^{6} - x^{2}$ $x^6+x^4y^2+y^4x^2+y^6-x^2$ , 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

$r^{2} = sin θ$ $r^2= sintheta$

Multiply both sides by $r$ $r$ :
$r^{2} \cdot r = r \cdot sin θ$ $r^2*r= r*sintheta$

Since $r sin θ = y$ $rsintheta= y$ and $r^{2} = x^{2} + y^{2}$ $r^2= x^2+y^2$ :

$(x^{2} + y^{2}) \cdot r = y$ $(x^2+y^2)*r= y$

And $r = \pm \sqrt{x^{2} + y^{2}}$ $r= +-sqrt(x^2+y^2)$

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

$(x^{2} + y^{2}) = \frac{y}{r}$ $(x^2+y^2)= y/r$

$(x^{2} + y^{2}) = \frac{y}{\pm \sqrt{x^{2} + y^{2}}}$ $(x^2+y^2)= y/(+-sqrt(x^2+y^2))$

Square both sides to see where this goes:

${(x^{2} + y^{2})}^{2} = {(\frac{y}{\pm \sqrt{x^{2} + y^{2}}})}^{2}$ $(x^2+y^2)^2= (y/(+-sqrt(x^2+y^2)))^2$

$x^{4} + 2 x^{2} y^{2} + y^{4} = \frac{y^{2}}{x^{2} + y^{2}}$ $x^4+2x^2y^2+y^4= y^2/(x^2+y^2)$

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

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

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]}