What is the Cartesian form of $r = -sin^2theta-r^2csc^2theta$ ?

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Answer 1 · 2017-01-03T09:08:58

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

The relation between polar coordinates $(r,theta)$ and Cartesian coordinates $(x,y)$ is

$x=rcostheta$ and $y=rsintheta$ i.e. $r^2=x^2+y^2$ and $y/x=tantheta$

Hence $r=-sin^2theta-r^2csc^2theta$ can be written as

$r=-y^2/r^2-r^2xxr^2/y^2=y^2/r^2-r^4/y^2$ and multiplying each term by $r^2y^2$ , this can be written as

$r^3y^2=-y^4-r^6$

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

What is the Cartesian form of r = -sin^2theta-r^2csc^2theta r = -sin^2theta-r^2csc^2theta ?