What is the Cartesian form of $(r+3)^2 = sin^3theta+csctheta$ ?

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Answer 1 · 2016-10-11T15:51:34

We know the relations

$x=rcostheta and y =rsintheta$

Again $x^2+y^2=r^2$

where r and $theta$ are the polar coordinate of a point having rectangular coordinate $(x,y)$

The given equation in polar form is

$(r+3)^2=sin^3theta+csctheta$

$=>r^2+6r+9=(r^3sin^3theta)/r^3+r/(rsintheta)$

$=>r^2+6r+9=y^3/r^3+r/y$

$=>yr^5+6yr^4+9yr^3=y^4+r^4$

$=>yr^5+(6y-1)r^4+9yr^3=y^4$

$=>y(x^2+y^2)^(5/2)+(6y-1)(x^2+y^2)^2+9y(x^2+y^2)^(3/2)=y^4$

This is the Cartesian form of the given polar equation.

What is the Cartesian form of (r+3)^2 = sin^3theta+csctheta (r+3)^2 = sin^3theta+csctheta ?