Double integrals with polar coordinates and center of mass.

[Daniel_Feldman]

1. Solved, thanks.

2. Consider the region inside x^2+y^2=6y and outside x^2+y^2=9. A thin lamina is formed from this region, and I need to find the lamina's center of mass if the density at any point is inversely proportional to its distance from the origin. Also, I need to explain why we do not care what the constant of proportionality is.

I drew the picture, but I'm not sure what the density function would be. Do I just take a point (x,y) and then density would be p(x,y)=1/sqrt(y^2-x^2)?

Any help would be great.

 

[galactus]

#1 converted to polar. Use \(\displaystyle x=rcos{\theta}\) and \(\displaystyle y=rsin{\theta}\)

Don't forget that extra r in polar. You're OK, except that should be r^5, not r^3.

Because, \(\displaystyle r\cdot{r^{2}}\cdot{r^{2}}=r^{5}\)


\(\displaystyle \L\\\int_{0}^{\pi}\int_{0}^{2}r(rcos{\theta})^{2}(rsin{\theta})^{2}drd{\theta}=\frac{4{\pi}}{3}\)

 

[Daniel_Feldman]

Oops...thanks.