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A component of a stone plinth has a square base with corners at coordinates $(0,0,0,)$, $(0,2,0)$, $(2,2,0)$ and $(2,0,0)$. The height of the top surface is defined as $z =(1 + x^2 y)$. (All units are in meters.)

If the density of the stone is $6000kg/m^3$ , calculate, to $2d.p$,

a. the mass of the stone block.

b. the x and y coordinates of its centre of mass.

Edit: What I know about solving this sort of problem is that usually you'd have a double integral of the surface density between the limits of the shape. However here since I am not given the surface density only the density and height I tried working out the mass of the object using the formula $p$ = $m$/$v$ where $p$ is the given density $6000kg/m^3$ and $v$ = $l$ * $w$ * $h$ if I assume that that the base can be treated as a cuboid. I then proceeded to integrate this to find the mass and the centre of mass of the shape. However I ran into problems when the mass was around $224000 kg$. The centre of mass using the method I found to be $(1.29 , 1.19)$ which seems reasonable as it lies within the area but I am not sure if my method is valid or the correct way to go about doing this. Apologies for not including this detail earlier, I was unaware of the amount of detail I had to go into.

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  • $\begingroup$ I know how to find mass and c.o.m if given a surface density but I'm not sure how to go about finding that from this. I tried working backwards from p =m/v where v would be the volume of a cuboid with height z = (1 + x^2 y) and sides length and width 2 but I'm not sure if that is the correct thing to do $\endgroup$ – user490418 Nov 30 '17 at 17:06
  • $\begingroup$ Yes of course. I am aware that I need a double integral both with limits 2 and 0. What I do not know is the function that I am supposed to integrate or how to get to it. $\endgroup$ – user490418 Nov 30 '17 at 17:14
  • $\begingroup$ Thank you for your answer, however, what you are suggesting is not helping me answer this question. If you could explain how for example the mass of the block is derived using the information I provided that would be more helpful, $\endgroup$ – user490418 Nov 30 '17 at 17:36
  • $\begingroup$ The density is the mass per unit volume. Therefore the total mass is obtained by integrating the (constant) density over the three dimensional volume, or alternatively as a double integral where you multiply the density by the height to simplify things. The center of mass is the point where all three moments (in $x,y,z$ directions) are zero. You don't convey that you've started with much understanding of the problem, so I'm not surprised to hear you complain explaining the problem to you "is not helping me answer this question." But we are trying to help you learn the material. $\endgroup$ – hardmath Nov 30 '17 at 23:20
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    $\begingroup$ Edit: I have tried what you suggested hardmath that is the double integral where I multiply the density by the height alone and I surprisingly got the same c.o.m that I talk about in my edit but with a different, much smaller, mass. I was wondering why this might be the case and which one would be correct. $\endgroup$ – user490418 Dec 1 '17 at 2:40
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I get the same (approximate) $x$-coordinate for the centroid below, but since the shape of the block is not a cuboid, my total mass for the block is different.

The general setup for centroid calculations (center of mass when the density is uniform in a body) is to compute "average" coordinates.

Here we were only asked to get the "average" $x$ and $y$ coordinates, but the average $z$ is defined in the same way. For convenience let's denote the constant density $\rho = 6000\; kg/m^3$. This constant affects the actual mass of the block (since mass = density * volume in compatible units), but it will cancel out in each fraction for a coordinate of the centroid.

Average $x$-coordinate:

$$ \overline{x} = \frac{\int_0^2 \int_0^2 \int_0^{x^2y+1} \rho x\; dz\; dy\; dx}{\int_0^2 \int_0^2 \int_0^{x^2y+1} \rho\; dz\; dy\; dx} $$

Average $y$-coordinate:

$$ \overline{y} = \frac{\int_0^2 \int_0^2 \int_0^{x^2y+1} \rho y\; dz\; dy\; dx}{\int_0^2 \int_0^2 \int_0^{x^2y+1} \rho\; dz\; dy\; dx} $$

Average $z$-coordinate:

$$ \overline{z} = \frac{\int_0^2 \int_0^2 \int_0^{x^2y+1} \rho z\; dz\; dy\; dx}{\int_0^2 \int_0^2 \int_0^{x^2y+1} \rho\; dz\; dy\; dx} $$

The denominator common to these expressions is the mass of the block, and the simplest of the integrals to evaluate:

$$ \begin{align*} m &= \int_0^2 \int_0^2 \int_0^{x^2y+1} \rho\; dz\; dy\; dx \\ &= \rho\; \int_0^2 \int_0^2 \int_0^{x^2y+1} dz\; dy\; dx \\ &= \rho\; \int_0^2 \int_0^2 (x^2y+1)\; dy\; dx \\ &= \rho\; \int_0^2 (2x^2+2)\; dx \\ &= \rho\; \left( \frac{16}{3}+4 \right) \\ &= \frac{28\rho}{3} \end{align*}$$

Thus a density of $6000\; kg/m^3$ times a volume of $28/3 \; m^3$ gives the block a mass of $56000\; kg$. Note that the triple integral uses the exact shape of the block (square base $z=0$ in $xy$-plane, but height $z$ of top surface varies with $x$ and $y$, $z=1+x^2y$ as specified).

When computing (say) the $x$-coordinate of the centroid, the density $\rho$ can be factored out and cancelled, and we can apply our previous calculation of volume:

$$ \begin{align*} \overline{x} &= \frac{\int_0^2 \int_0^2 \int_0^{x^2y+1} \rho x\; dz\; dy\; dx}{\int_0^2 \int_0^2 \int_0^{x^2y+1} \rho\; dz\; dy\; dx} \\ &= \frac{\rho\;\int_0^2 \int_0^2 \int_0^{x^2y+1} x\; dz\; dy\; dx}{\rho\;\int_0^2 \int_0^2 \int_0^{x^2y+1} dz\; dy\; dx} \\ &= \frac{\int_0^2 \int_0^2 \int_0^{x^2y+1} x\; dz\; dy\; dx}{\int_0^2 \int_0^2 \int_0^{x^2y+1} dz\; dy\; dx} \\ &= \left( \frac{3}{28}\; \right) \int_0^2 \int_0^2 \int_0^{x^2y+1} x\; dz\; dy\; dx \\ &= \left( \frac{3}{28}\; \right) \int_0^2 \int_0^2 x(x^2y+1)\; dy\; dx \\ &= \left( \frac{3}{28}\; \right) \int_0^2 (2x^3+2x)\; dx \\ &= \left( \frac{3}{28}\; \right) 12 \approx 1.29 \end{align*} $$

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