How do you convert $- y = 3 y^{2} - x^{2} + x^{3}$ $-y=3y^2-x^2 +x^3$ into a polar equation?

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Answer 1 · 2016-04-15T20:58:44

$0 = 3 r^{2} {sin}^{2} θ + r sin θ + r^{3} {cos}^{3} θ - r^{2} {cos}^{2} θ$ $0=3r^2sin^2theta+rsintheta+r^3cos^3theta-r^2cos^2theta$

$0 = 3 y^{2} + y + x^{3} - x^{2}$ $0=3y^2+y+x^3-x^2$
use the formulas:
$x = r cos θ and y = r sin θ$ $x=rcostheta and y= rsin theta$

$0 = 3 r^{2} {sin}^{2} θ + r sin θ + r^{3} {cos}^{3} θ - r^{2} {cos}^{2} θ$ $0=3r^2sin^2theta+rsintheta+r^3cos^3theta-r^2cos^2theta$

How do you convert -y=3y^2-x^2 +x^3 −y=3y2−x2+x3-y=3y^2-x^2 +x^3 into a polar equation?