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I am trying to find the $y$-coordinate of the center of mass of a semi-annular plane region bounded by $1\leq x^2+y^2 \leq 4$ and $ y > 0$. I know that the sketch should look similar to half a donut above the y axis, but I am not sure how to set up the integral with the appropriate limits.

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3 Answers 3

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Assuming that the density of the region is uniform, the $y$ coordinate of the center of mass is given as $$y_c = \dfrac{\int_R y dx dy}{\int_R dx dy}$$

The region, $R$, you are interested in i.e. the semi-annular plane as shown below. enter image description here Converting this to polar coordinates, note that the $r$ varies from $1$ to $2$ while $\theta$ varies from $0$ to $\pi$. Now you should be able to the integral.

Move your mouse over the gray area for a hint.

Converting to polar coordinates, we get that $y = r \sin(\theta)$ and the area element $dxdy$ gets converted to $r dr d \theta$. Hence, this gives us the center of mass as $$y_c = \dfrac{\int_{r=1}^{2} \int_{\theta=0}^{\pi} r \sin(\theta) r dr d \theta}{\int_{r=1}^{2} \int_{\theta=0}^{\pi} r dr d \theta}$$

Move your mouse over the gray area below for the complete solution.

Hence, we get that \begin{align}y_c & = \dfrac{\int_{r=1}^{2} \int_{\theta=0}^{\pi} r \sin(\theta) r dr d \theta}{\int_{r=1}^{2} \int_{\theta=0}^{\pi} r dr d \theta} = \dfrac{\left(\int_{r=1}^{2} r^2 dr \right) \left(\int_{\theta=0}^{\pi} \sin(\theta) d \theta \right)}{\left(\int_{r=1}^{2} r dr \right) \left(\int_{\theta=0}^{\pi} d \theta \right)}\\ & = \dfrac{\left(\dfrac{2^3-1^3}{3} \right) \times2}{\left(\dfrac{2^2-1^2}{2} \right)\times \pi} = \dfrac{14}{3} \times \dfrac{2}{3 \pi} = \dfrac{28}{9 \pi}\end{align}

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The method for converting to polar coordinates for r is not obvious to me because I don't see how you found the inner radius $r=1$ of the annulus... was this determined by the lower bound, i.e. $1 \leq x^2 + y^2 \leq 4$? –  Dylan Jun 12 '12 at 6:25
    
@Dylan In polar coordinates, $x = r \cos(\theta)$ and $y = r \sin(\theta)$. Hence, $x^2 + y^2 = r^2$. This means $1 \leq r^2 \leq 4 \implies 1 \leq r \leq 2$. –  user17762 Jun 12 '12 at 6:28
    
I see. That makes sense now. Thank you. –  Dylan Jun 12 '12 at 6:34

We want to find the moment of the region about the $x$-axis, and divide by the area of the region. The area is no problem, it is basic geometry, so we deal with the moment. (But we must not forget to divide at the end!)

The easiest way is to note that the moment is the moment of the upper half of the circle $x^2+y^2=4$, minus the moment of the upper half of the circle $x^2+y^2=1$.

The two moment calculations are essentially the same. For the big half-circle the moment is $$\int_{-2}^2 \frac{1}{2}y^2\,dx.$$ Replace $y^2$ by $4-x^2$, and integrate. Now do the same calculation for the small half-circle, and subtract.

Why? Take a vertical strip, running from $x$ to $x+dx$. We can think of the mass as concentrated halfway up, at $y=\frac{1}{2}\sqrt{4-x^2}$. The mass is roughly $y\,dx$, that is $\sqrt{4-x^2}\,dx$. Multiply the mass by its effective distance from the $x$-axis. We get $(y\,dx)(\frac{1}{2}y)$. Now replace $y^2$ by $4-x^2$, and "add up" (integrate) from the beginning $-2$ to the end $2$.

We can save ourselves some trouble with minus signs by noting that the height of the centre of mass is the height of the centre of mass of the right half of the region.

Alternately, we can try to deal with the whole region between the circles directly. Then we sort of run into setup issues that may have caused you trouble. The problem is that the region looks and is different from $x=-1$ to $x=1$ than it is from $x=1$ to $x=2$, or from $-2$ to $-1$. So we will have to break up the region of integration.

From $-1$ to $1$, we have a mass of $(\sqrt{4-x^2}-\sqrt{1-x^2})\,dx$ concentrated halfway between $\sqrt{4-x^2}$ and $\sqrt{1-x^2}$, that is, at $\frac{1}{2}(\sqrt{4-x^2}+\sqrt{1-x^2})$. So we find $$\int_{-1}^1 \frac{1}{2}(\sqrt{4-x^2}+\sqrt{1-x^2})(\sqrt{4-x^2}-\sqrt{1-x^2})\,dx.$$ The integrand simplifies very nicely to $3$, so the integration is trivial.

For the two end pieces, we do a similar calculation, and add the results to the number just obtained. We deal with the end piece in the first quadrant. The moment is $$\int_1^2 \frac{1}{2}\sqrt{4-x^2}\sqrt{4-x^2}\,dx.$$ Things simplify nicely, we are integrating $\frac{1}{2}(4-x^2)$, and the calculation is easy. We also need to add the moment of the piece between $-2$ and $-1$, but by symmetry that is the same.

Note that the "alternate" calculation, though not hard, certainly takes more writing.

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This solution seems a lot less involved then what I initially thought... I thought that finding the moment involved a double integral such that $\int \int \rho f(x,y)$ –  Dylan Jun 12 '12 at 4:44
    
@Dylan: If we have a piece of planar material with variable density $\rho(x,y)$, then the multiple integral can in general not be avoided. But the problem you posed, which is the geometric moment for constant density, which we can take to be $1$, is fundamentally a $1$ variable problem. Of course we need also the $x$-coordinate of the centre of mass, but you did not ask about that, recognizing that by symmetry this $x$-coordinate is $0$. –  André Nicolas Jun 12 '12 at 5:03

Use Pappus-Guldin theorem: http://mathworld.wolfram.com/PappussCentroidTheorem.html. Namely: if you rotate the are in your figure, the volume it generates = area_surface x 2$\pi$ x h (h = the sought height of the centre of mass of that area).

We have: $\frac{4\pi}{3}(2^3-1^3) = 2\pi h \frac{\pi(2^2-1^2)}{2} \;\;$. The result is $h=\frac{28}{9\pi}$.

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