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The motion of a solid object can be analyzed by thinking of the mass as concentrated at a single point, the center of mass. If the object has density enter image description here at the point enter image description here and occupies a region W, then the coordinates enter image description here of the center of mass are given by enter image description here where enter image description here is the total mass of the body. Consider a solid is bounded below by the square enter image description here ,enter image description here, enter image description here and above by the surfaceenter image description here the density of the solid be 1 g/cm$^3$, with x,y,z measured in cm. Find each of the following:

The mass of the solid

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I am having troubles with starting and setting the integral

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  • 2
    $\begingroup$ FYI the "motion of a solid object..." is just color commentary which is irrelevant to the question you're posing. $\endgroup$ – Jonathan Mar 25 '14 at 2:48
  • $\begingroup$ Also because in a real rigid motion you have to consider rotational momenta (which gives you the phase space $T^*(\mathbb{R}^3\times SO(3))$) and not only the center of mass ($T^*\mathbb{R}^3$). $\endgroup$ – Daniel Robert-Nicoud Sep 30 '15 at 5:17
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  1. The mass of the solid is $\int_W \rho \mathrm{d}V$ which in this case is equal to $\rho \int_W \mathrm{d}V$ as the solid has a uniform mass density. To find the mass consider a small rectangular element of sides $\mathrm{d}x$ and $\mathrm{d}y$ on the $xy$ plane. The volume of the cuboidal rod of solid with this element as the base is $z\mathrm{d}x\mathrm{d}y$. So the total volume of the solid will be $$\int_0^5\int_0^4(x+y+3) \mathrm{d}x \mathrm{d}y = 150$$ and therefore the mass = $1gm/cm^3\cdot150 cm^3 = 150 gm$

  2. For the coordinates of center of mass the integrals are similar $$ \overline{x} = \frac{1}{m}\int_0^5\int_0^4\rho x(x+y+3) \mathrm{d}x \mathrm{d}y$$ $$ \overline{y} = \frac{1}{m}\int_0^5\int_0^4\rho y(x+y+3) \mathrm{d}x \mathrm{d}y$$ $$ \overline{z} = \frac{1}{m}\int_0^5\int_0^4\rho \frac{(x+y+3)^2}{2} \mathrm{d}x \mathrm{d}y$$

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The mass is

$$ m = \int_0^4 \int_0^5 \int_0^{x+y+3}\,\rho \mathrm{d}z\mathrm{d}y\mathrm{d}x = 150 \rho $$

The center of mass is

$$\begin{pmatrix} \overline{x} \\ \overline{y} \\ \overline{z} \end{pmatrix} = \frac{1}{m} \int_0^4 \int_0^5 \int_0^{x+y+3}\,\rho \begin{pmatrix}x\\y\\z\end{pmatrix}\mathrm{d}z\mathrm{d}y\mathrm{d}x = \frac{1}{150 \rho} \begin{pmatrix} \frac{980 \rho}{3} \\ \frac{1250 \rho}{3} \\ \frac{1790 \rho}{3} \end{pmatrix} = \ldots$$

This is how you setup the integrals.

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