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April 30th, 2017, 10:33 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  Triple Integral of a volume of a pyramid
Find the volume of the pyramid with corners at (1,1, 0), (1,1, 0), (1, 1, 0), (1, 1, 0), and (0, 0, 2) $\displaystyle \int_{1}^{1}\int_{1}^{1}\int_{0}^{??????} dz dy dx$ What would be the equation of upper limit z (height) in relation to x and y? 
April 30th, 2017, 11:45 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 
you need to learn to see the differential volumes in this case you've got squares of infinitesimal thickness in the z direction each square has side length $s=2z$ the differential volume of these is $dV = s^2 ~dz = (2z)^2~dz$ $V = \displaystyle \int_0^2~(2z)^2~dz = \dfrac 8 3$ 
April 30th, 2017, 12:44 PM  #3  
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  Quote:
 
April 30th, 2017, 12:50 PM  #4 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
wait... I don't understand why each square has side length s=2−z

April 30th, 2017, 02:21 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372  
May 1st, 2017, 03:04 PM  #6 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
I need help.... The original question is actually as following: An object occupies the volume of the pyramid with corners at (1, 1, 0), (1, 1, 0), (1, 1, 0), (1, 1, 0), and (0, 0, 2) and has density x^2 + y^2 at (x, y, z). Find the center of mass. Answer: (0, 0, 1/3) This is as far as I could come up: $\displaystyle \int_{1}^{1}\int_{1}^{1}\int_{0}^{??????} dz dy dx$ 
May 2nd, 2017, 11:34 AM  #7  
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372  Quote:
So here's a version with a relatively straightforward shortcut. Consider the infinitesimal volume that is the pyramid cut by a plane parallel to the $xy$ plane that intersects the $z$axis at $z$ This will be a square with corners $\left(1+\frac z 2,1+\frac z 2\right),\left(1+\frac z 2,1\frac z 2\right),\left(1\frac z 2,1\frac z 2\right),\left(1\frac z 2,1+\frac z 2\right)$ and differential thickness $dz$ The differential mass of this differential volume is given by $dM = \displaystyle \left(\int_{1+\frac z 2}^{1\frac z 2}\int_{1+\frac z 2}^{1\frac z 2}~x^2+y^2~dy~dx\right)~dz$ I'll let you sweat the details but $dM = \dfrac{1}{6} (z2)^4~dz$ and now $M=\displaystyle \int_0^2~dM = \dfrac{16}{15}$ Now we note that with the symmetry of both the density and the pyramid itself that the centroid will be on the $z$axis so we need only find the $z$ direction moment. Because the density is independent of $z$ we can use $dM$ above to obtain $mom_z = \displaystyle \int_0^2 z dM = \int_0^2 \dfrac{z}{6} (z2)^4 ~dz = \dfrac{16}{45}$ and the centroid will be at $C = \left(0,0,\dfrac{mom_z}{M}\right ) = \left( 0,0,\dfrac{ \frac{16}{45}}{ \frac{16}{15}} \right) = \left(0,0,\dfrac 1 3 \right)$  
May 2nd, 2017, 02:11 PM  #8 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
What is your intuition behind z/2?

May 2nd, 2017, 02:38 PM  #9 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372  

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