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A lamina occupies the region which is the intersection of $x^2+y^2-2y \leq 0$ and the first quadrant of the $xy$-plane. Find the center of mass if the density at a point of the lamina is twice the point's distance from the origin.

Does the setup of this look like this:$$\int_0^{\frac{\pi}{2}}\int_0^{2\sin\theta}2r^2drd\theta?$$

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I'm assuming $(x, y) \in \mathbb{R}^2$. If so, then $\{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \leq 0\} = \{(0, 0)\}$. –  Michael Albanese Apr 18 '13 at 16:15
    
Sorry, forgot the (-2y), you have to complete the square so the radius becomes 1. –  user73064 Apr 18 '13 at 16:18

1 Answer 1

Actually, the factor of $2$ is unimportant. The center of mass in the $y$ direction is given by

$$\bar{y} = \frac{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)(r \sin{\theta})}{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)}$$

Note that the weight of the $y$ coordinate is expressed in the $r \sin{\theta}$ term in the numerator. Note also the boundary of the lamina is expressed in its polar form, $r=\sin{\theta}$. Evaluating the radial integrals, we get

$$\bar{y} = \frac{\frac{16}{4} \displaystyle \int_0^{\pi/2} d\theta \: \sin^5{\theta}}{\frac{8}{3} \displaystyle \int_0^{\pi/2} d\theta \: \sin^3{\theta}}$$

or

$$\bar{y} = \frac{6}{5}$$

For $\bar{x}$, we do a similar calculation:

$$\begin{align}\bar{x} &= \frac{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)(r \cos{\theta})}{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r (2 r)}\\ &= \frac{\frac{16}{4} \displaystyle \int_0^{\pi/2} d\theta \: \cos{\theta} \sin^4{\theta}}{\frac{8}{3} \displaystyle \int_0^{\pi/2} d\theta \: \sin^3{\theta}}\\ &= \frac{9}{20}\end{align}$$

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I originally meant the equation for the mass because once you have that it's very easy to go from mass to center of mass...I don't understand why you went from (0 to pi) and(0 to sinθ)...if the lamina is restricted to the first quadrant, shouldn't it be (0 to pi/2)? –  user73064 Apr 18 '13 at 17:23
    
I didn't catch the first quadrant, let me fix. That will not affect the $y$ result, but will affect the $x$. –  Ron Gordon Apr 18 '13 at 17:23
    
Okay, I still don't get how you can just get rid of the 2 in r=2sinθ? It would result in a different answer so I don't see how it's unimportant. –  user73064 Apr 18 '13 at 17:38
    
Because the center of mass calculation involves a ratio - the denominator is the total mass. In doing that calculation, any constant factor will cancel with the numerator. The important thing is that the density is proportional to $r$. –  Ron Gordon Apr 18 '13 at 17:39
    
But it wouldn't be wrong to put 0 to 2sinθ as a bound? –  user73064 Apr 18 '13 at 17:59

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