How do you convert #r = 3sin(theta)# into a cartesian equation?

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Answer 1 · 2015-09-22T20:21:36

Use #r = sqrt(x^2+y^2)# and #sin(theta) = y / r# to get:

#x^2+y^2 = 3y#

#r = sqrt(x^2+y^2)# and #sin(theta) = y / r#

So #r = 3sin(theta)# becomes:

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

Multiply both sides by #sqrt(x^2+y^2)# to get:

#x^2+y^2 = 3y#

graph{x^2+y^2-3y=0 [-4.913, 5.087, -0.98, 4.02]}