Question

Calculus 3 - Applications of Double Integrals

The formula to find the centroid is given by
x̄ = (1/m) ∫∫xρ(x,y)dA
Y-bar = (1/m) ∫∫yρ(x,y)dA

If I'm given a polar equation, such as r=sinθ, how would I proceed to find the centroid? Is there also a centroid formula for the polar equation or do I have to convert the polar equation into a cartesian equation?

Answer 1

Sorry I am not of more help but in my own personal work, I do not believe there is a centroid formula for the polar equation.

Answer 2

idk if there is polar formula for centroid.
but you can always do any of these 2 things
1) as you said, convert the equation into cartesian and use the formula
r = sin t , r^2 = r sin t , gives x^2+y^2 = y and so on

or 2) you can try to convert the formula altogether
x = r cos t
dA = r dr dt

x̄ = (1/m) ∫∫xρ(x,y)dA changes to
x̄ = (1/m) ∫∫ r cos t ρ(r,t) (r dr dt)