Consider a halved hard torus (half a donut). The radius the the torus space $R_1$ and also $R_2$. I need to uncover its center of mass. The note they give is that the center of fixed of a homogeneous solid object $\Omega \subset \Bbb R^3$ is calculate as$$\overlinex=\int_\Omega\overlinerd\overliner.$$

I yes, really don"t recognize this formula, ns don"t understand what $\overliner$ means and what is $\Omega$ in this case. I"d appreciate that someone explains what this formula method and exactly how to apply it in this problem. Many thanks in advance.

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asked Oct 19 "13 in ~ 0:01

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I discovered it most basic to use cylindrical coordinates to set up the integrals essential for the facility of mass. Before we execute so, however, I set my coordinate device up as follows. I have positive $x$ coming the end of the screen, optimistic $y$ going come the right, and also positive $z$ up. In cylindrical coordinates $(r,\phi,z)$:

$$x = r \cos\phi$$$$y=r \sin\phi$$

where we have the boundaries defining the an ar $\Omega$:

$$\phi \in \left < -\frac\pi2,\frac\pi2 \right>$$$$z \in <-R_2,R_2>$$$$r \in \left < R_1 - \sqrtR_2^2-z^2, R_1 + \sqrtR_2^2-z^2\right>$$

Also, for an object of consistent mass density, the expression because that the $x$ ingredient of the facility of mass is

$$\barx = \frac\displaystyle \int_\Omega d^3 \vecx \, x\displaystyle \int_\Omega d^3 \vecx$$

(Note that, by symmetry, we have actually $\bary=0$ and $\barz=0$.) Let"s an initial evaluate the denominator:

\beginalign \int_\Omega d^3 \vecx &= \int_-\pi/2^\pi/2 d\phi \, \int_-R_2^R_2 dz \, \int_R_1 - \sqrtR_2^2-z^2^R_1 + \sqrtR_2^2-z^2 dr \, r \\ &= \frac\pi2\int_-R_2^R_2 dz \left <\left (R_1 + \sqrtR_2^2-z^2 \right )^2 - \left (R_1 - \sqrtR_2^2-z^2 \right )^2 \right >\\ &= 4 \pi R_1 \int_0^R_2 dz \, \sqrtR_2^2-z^2\\ &= \pi^2 R_1 R_2^2 \endalign

Now we evaluate the facility of mass:

\beginalign\barx &= \frac1\pi^2 R_1 R_2^2 \int_-\pi/2^\pi/2 d\phi \, \int_-R_2^R_2 dz \, \int_R_1 - \sqrtR_2^2-z^2^R_1 + \sqrtR_2^2-z^2 dr \, r^2 \cos\phi \\ &= \frac43 \pi^2 R_1 R_2^2 \int_0^R_2 dz \, \left <\left (R_1 + \sqrtR_2^2-z^2 \right )^3 - \left (R_1 - \sqrtR_2^2-z^2 \right )^3 \right >\\ &= \frac83 \pi^2 R_1 R_2^2 \int_0^R_2 dz \,\left <3 R_1^2 \sqrtR_2^2-z^2 + \left (R_2^2-z^2 \right )^3/2 \right > \\ &= \frac83 \pi^2 R_1 R_2 \left ( \frac3 \pi4 R_1^2 R_2 + \frac3 \pi16 R_2^3 \right )\endalign

Simplifying, i get

$$\barx = \frac4 R_1^2+R_2^22 \pi R_1$$

As a fast note, in the borders as $R_2 \to 0$, we discover that the facility of fixed becomes