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]}