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April 30th, 2017, 10:33 AM   #1
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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?
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April 30th, 2017, 11:45 AM   #2
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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 $s=2-z$

the differential volume of these is

$dV = s^2 ~dz = (2-z)^2~dz$

$V = \displaystyle \int_0^2~(2-z)^2~dz = \dfrac 8 3$
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April 30th, 2017, 12:44 PM   #3
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Quote:
Originally Posted by romsek View Post
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 $s=2-z$

the differential volume of these is

$dV = s^2 ~dz = (2-z)^2~dz$

$V = \displaystyle \int_0^2~(2-z)^2~dz = \dfrac 8 3$
Absolutely amazing... I need to contemplate about your suggestion of the differential volumes.
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April 30th, 2017, 12:50 PM   #4
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wait... I don't understand why each square has side length s=2−z
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April 30th, 2017, 02:21 PM   #5
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Quote:
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wait... I don't understand why each square has side length s=2−z
in the xy plane the length is 2. Look at the corner points.

at $z=2$ the square is just a point, i.e. $s=0$

The relation is linear so it must be $s=2-z$
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May 1st, 2017, 03:04 PM   #6
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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:
$\displaystyle \int_{-1}^{1}\int_{-1}^{1}\int_{0}^{??????} dz dy dx$
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May 2nd, 2017, 11:34 AM   #7
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Quote:
Originally Posted by zollen View Post
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:
$\displaystyle \int_{-1}^{1}\int_{-1}^{1}\int_{0}^{??????} dz dy dx$
ok... I confess I tried solving this w/o any shortcuts and I couldn't make the numbers work out.

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

$dM = \dfrac{1}{6} (z-2)^4~dz$

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

$mom_z = \displaystyle \int_0^2 z dM = \int_0^2 \dfrac{z}{6} (z-2)^4 ~dz = \dfrac{16}{45}$

and the centroid will be at

$C = \left(0,0,\dfrac{mom_z}{M}\right ) = \left( 0,0,\dfrac{ \frac{16}{45}}{ \frac{16}{15}} \right) = \left(0,0,\dfrac 1 3 \right)$
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May 2nd, 2017, 02:11 PM   #8
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What is your intuition behind z/2?
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May 2nd, 2017, 02:38 PM   #9
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Quote:
Originally Posted by zollen View Post
What is your intuition behind z/2?
a) linear

b) z=2 @x=0

c) z=0 @x=1, -1

z = 2-2|x|

x = 1-z/2, -1+z/2

all this applies to y as well as x
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