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Homework Statement:
 Find the centre of mass of a uniform sector of radius a and angle θ.
Relevant Equations:

##A=\frac{1}{2}a^2\theta##
Average value of value of ##f## over ##A=\frac{1}{A}\iint f\, dA##
If I use cartesian coordinates, I get:
##\bar{x}=\frac{1}{A}\iint x\, dA=\frac{1}{A} \iint r^2\cos\theta\, dr\, d\theta= \frac{2a\sin\theta}{3\theta}##
##\bar{y}=\frac{1}{A}\iint y\, dA=\frac{1}{A}\iint r^2\sin\theta\, dr\, d\theta= \frac{2a(1\cos\theta)}{3\theta}##
But if I use polar coordinates, I get:
##\bar{r}=\frac{1}{A}\iint r\,dA=\frac{1}{A}\iint r^2\, dr\, d\theta=\frac{2a}{3}##
##\bar{\theta}=\frac{1}{A}\iint \theta\,dA=\frac{1}{A}\iint r\theta\, dr\, d\theta=\frac{\theta}{2}##
I believe the cartesian coordinates are correct and obviously the value of ##\bar{\theta}## is correct, but the polar coordinates of the centroid do not correspond with the cartesian coordinates. What is wrong with my calculation of ##\bar{r}##?
##\bar{x}=\frac{1}{A}\iint x\, dA=\frac{1}{A} \iint r^2\cos\theta\, dr\, d\theta= \frac{2a\sin\theta}{3\theta}##
##\bar{y}=\frac{1}{A}\iint y\, dA=\frac{1}{A}\iint r^2\sin\theta\, dr\, d\theta= \frac{2a(1\cos\theta)}{3\theta}##
But if I use polar coordinates, I get:
##\bar{r}=\frac{1}{A}\iint r\,dA=\frac{1}{A}\iint r^2\, dr\, d\theta=\frac{2a}{3}##
##\bar{\theta}=\frac{1}{A}\iint \theta\,dA=\frac{1}{A}\iint r\theta\, dr\, d\theta=\frac{\theta}{2}##
I believe the cartesian coordinates are correct and obviously the value of ##\bar{\theta}## is correct, but the polar coordinates of the centroid do not correspond with the cartesian coordinates. What is wrong with my calculation of ##\bar{r}##?
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