How do you convert #r^2 = 4 sin θ# into a cartesian equation?

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Answer 1 · 2015-10-21T17:09:41

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

Given that

#x^2+y^2=r^2#

#y=rsin\theta#

#sin\theta=y/r#

#r=sqrt(x^2+y^2)#

We proceed as follows

#x^2+y^2=4(y/r)#

#x^2+y^2=(4y)/sqrt(x^2+y^2)#

#(x^2+y^2)(x^2+y^2)^(1/2)=4y#

#(x^2+y^2)^(2/2)(x^2+y^2)^(1/2)=4y#

#(x^2+y^2)^(3/2)=4y#

NOTE: If we want that part of the graph below the x axis it is

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

Which corresponds to #r# going in the other direction in polar coordinates

We can check by converting back and see if we get what we started with

#(r^2)^(3/2)=4rsin\theta#

#r^3=4rsin\theta#

Dividing both sides by #r#

#r^2=4sin\theta#