mass of tetrahedron

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Question statement

Cosider the tetrahedron bounded by the coordinate planes $x=0$ , $y=0$ and $z=0$ and the plane $\frac{x}{\var{a}} + \frac{y}{\var{b}} + \frac{z}{\var{c}} =1$ . Assume that it has a density function $\rho(x) = \var{R0}(\var{a}+x)$ .

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			Name	Type	Generated Value

a	integer	1
b	integer	5
c	integer	2
r0	integer	1
Mass	number	2.0833333333
massCentrex	number	0.28
massCentrey	number	1.2
massCentrez	number	0.096

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Generated value: `integer`

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Mass
massCentrex
massCentrey
massCentrez

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The coordinate surfaces $x=0$ , $y=0$ and $z=0$ are given. The plane $\frac{x}{\var{a}} + \frac{y}{\var{b}} + \frac{z}{\var{c}} = 1$ can be rewritten as a functions as follows

$z = f(x,y) = \var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right).$

The surface of $f(x,y)$ intersects the axes at $x=\var{a}, y=\var{b}$ and $z=\var{c}$ .

We can write the boundaries of the tetrahedron as

$0<x<\var{a} \quad \mbox{and} \quad 0 < y< \frac{\var{b}}{\var{a}}(\var{a} -x) \quad\mbox{and }\quad 0\leq z \leq \var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right).$

Thus, with mass element $dM = \rho(x)dV = \rho(x)dx\,dy\,dz$ , the total mass is

$M= \int_T\,dM = \int_0^\var{a}\int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)}\int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)}\rho(x)\,dz\,dy\,dx =\\[3mm] \int_0^\var{a}\int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)}\int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)} \var{R0}(\var{a}+x) \,dz\,dy\,dx =\\[3mm] \int_0^\var{a} \var{R0}(\var{a}+x) \int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)}\int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)} \,dz\,dy\,dx =\\[3mm] \int_0^\var{a} \var{R0}(\var{a}+x) \int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)} \var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right) \,dy\,dx =\\[3mm] \int_0^\var{a} \var{R0}(\var{a}+x)\left[ \frac{\simplify{{b*c}}}{\simplify{{2*a^2}}}(\var{a}-x)^2\right]\, dx = \frac{5\cdot\var{a}^2\cdot\var{b}\cdot\var{c}\cdot\var{R0}}{24}=\frac{\simplify{{5*{a^2}*{b}*{c}*{R0}}}}{24}.$

We start with the $x$ coordinate.

$\bar x= \frac{1}{M}\int_T x\,dM = \int_0^\var{a}\int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)}\int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)}x\rho(x)\,dz\,dy\,dx =\\[3mm] \frac{1}{M} \int_0^\var{a}\int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)}\int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)} x\var{R0}(\var{a}+x) \,dz\,dy\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} x\var{R0}(\var{a}+x) \int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)}\int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)} \,dz\,dy\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} x\var{R0}(\var{a}+x) \int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)} \var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right) \,dy\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} x\var{R0}(\var{a}+x)\left[ \frac{\simplify{{b*c}}}{\simplify{{2*a^2}}}(\var{a}-x)^2\right]\, dx =\\[3mm] \frac{\simplify{{R0*b*c}}}{\simplify{{2*a^2}}M}\int_0^\var{a} x (\var{a}+x)\left[ (\var{a}-x)^2\right]\, dx = \frac{\simplify{{R0*b*c}}}{\simplify{{2*a^2}}M}\int_0^\var{a}x(\var{a^3} -\var{a^2}x-\var{a} x^2 + x^3) \, dx =\\[3mm] \frac{\simplify{{R0*b*c}}}{\simplify{{2*a^2}}M}\int_0^\var{a}\var{a^3}x -\var{a^2}x^2-\var{a} x^3 + x^4 \, dx = \\[3mm] \frac{\simplify{{R0*b*c}}}{\simplify{{2*a^2}}M}\left[\frac{\var{a^3}x^2}{2} -\frac{\var{a^2}x^3}{3}-\frac{\var{a} x^4}{4} + \frac{x^5}{5} \right]_0^{\var{a}}=\\[3mm] \simplify[fractionNumbers]{({a}^5/2 - ({a}^5)/3 - {a}^5/4 + {a}^5/5) *{R0*b*c}/{2*a^2*Mass}} = \simplify[fractionNumbers,otherNumbers]{({a}^5/2 - ({a}^5)/3 - {a}^5/4 + {a}^5/5) *{R0*b*c}/{2*a^2*Mass}}.$

Simlarly, for $y$ coordinate:

$\bar y = \frac{1}{M}\int_T y\,dM = \int_0^\var{a}\int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)}\int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)}y\rho(x)\,dz\,dy\,dx =\\[3mm] \frac{1}{M} \int_0^\var{a}\int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)}\int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)} y\var{R0}(\var{a}+x) \,dz\,dy\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} \var{R0}(\var{a}+x) \int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)} y \int_0^{\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right)} \,dz\,dy\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} \var{R0}(\var{a}+x) \int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)} y\var{c}\left( 1 - \frac{x}{\var{a}} - \frac{y}{\var{b}}\right) \,dy\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} \var{R0}(\var{a}+x) \int_0^{\frac{\var{b}}{\var{a}}(\var{a} -x)} \var{c}\left( y - \frac{xy}{\var{a}} - \frac{y^2}{\var{b}}\right) \,dy\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} \var{R0}(\var{a}+x)\frac{\simplify{{c*b^2}}}{\simplify{{6*a^3}}}\left[\var{a^3}-3\cdot\simplify[unitFactor]{{a^2}x} +3\cdot\simplify[unitFactor]{{a}x^2}-x^3\right]\, dx =\\[3mm] \frac{\simplify{{R0*b^2*c}}}{\simplify{{6*a^3}}M}\int_0^\var{a} (\var{a}+x)( \var{a}-x)^3, dx = \frac{\simplify{{R0*b^2*c}}}{\simplify{{6*a^3}}M}\int_0^\var{a}\var{a^4} -2\cdot\simplify[unitFactor]{{a^3}x} +2\cdot\simplify[unitFactor]{{a}x^3}-x^4 \, dx =\\[3mm] \frac{\simplify{{R0*b^2*c}}}{\simplify{{6*a^3}}M}\left[\simplify[unitFactor]{{a^4}x} - \simplify[unitFactor]{{a^3}x^2} +\frac{\simplify[unitFactor]{{a}x^4}}{2}-\frac{x^5}{5} \right]_0^\var{a} = \\[3mm] \simplify[fractionNumbers]{({a}^5 - ({a}^5) + {a}^5 /2- {a}^5/5) *{R0*b^2*c}/{6*a^3*Mass}} = \simplify[fractionNumbers,otherNumbers]{({a}^5 - ({a}^5) + {a}^5 /2- {a}^5/5) *{R0*b^2*c}/{6*a^3*Mass}}.$

Finaly, for $z$ coordinate:

$\bar z= \frac{1}{M}\int_T z\,dM = \int_0^\var{a}\int_0^{\frac{\var{c}}{\var{a}}(\var{a} -x)}\int_0^{\var{b}\left( 1 - \frac{x}{\var{a}} - \frac{z}{\var{c}}\right)}z\rho(x)\,dy\,dz\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a}\int_0^{\frac{\var{c}}{\var{a}}(\var{a} -x)}\int_0^{\var{b}\left( 1 - \frac{x}{\var{a}} - \frac{z}{\var{c}}\right)} z\var{R0}(\var{a}+x) \,dy\,dz\,dx =\\[3mm] \frac{1}{M} \int_0^\var{a}\var{R0}(\var{a}+x)\int_0^{\frac{\var{c}}{\var{a}}(\var{a} -x)}z\int_0^{\var{b}\left( 1 - \frac{x}{\var{a}} - \frac{z}{\var{c}}\right)} \,dy\,dz\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} \var{R0}(\var{a}+x)\int_0^{\frac{\var{c}}{\var{a}}(\var{a} -x)} z\var{b}\left( 1 - \frac{x}{\var{a}} - \frac{z}{\var{c}}\right)\,dz\,dx =\\[3mm] \frac{1}{M}\int_0^\var{a} \var{R0}(\var{a}+x)\int_0^{\frac{\var{c}}{\var{a}}(\var{a} -x)} \var{b}\left( z - \frac{xz}{\var{a}} - \frac{z^2}{\var{c}}\right)\,dz\,dx =\\[3mm] \frac{\simplify{{R0*a}}}{M}\int_0^\var{a} (\var{a}+x) \left[ \frac{z^2}{2} - \frac{xz^2}{\var{2*a}} - \frac{z^3}{\var{3*c}}\right]_0^{\frac{\var{c}}{\var{a}}(\var{a} -x)}\,dx =\\[3mm] \frac{\simplify{{R0*c^2}}}{\simplify{{6*a^2}}M}\int_0^\var{a} (\var{a}+x)(a-x)^3\,dx = \frac{\simplify{{R0*c^2}}}{\simplify{{6*a^2}}M}\int_0^\var{a} \var{a^4}-\var{2*a^3}x + \var{2*a}x^3 - x^4\,dx\\[3mm] \frac{\simplify{{R0*c^2}}}{\simplify{{6*a^2}}M}\left[\var{a^4}x -\frac{\var{2*a^3}x^2}{2} + \frac{\var{a}x^4}{2} - \frac{x^5}{5}\right]_0^{\var{a}}=\\[3mm] \simplify[fractionNumbers]{({a^5} -{2*a^5}/2 + {a^5}/2 - {a^5}/5) *{R0*c^2}/{6*a^2*Mass}} = \simplify[fractionNumbers,otherNumbers]{({a^5} -{2*a^5}/2 + {a^5}/2 - {a^5}/5) *{R0*c^2}/{6*a^2*Mass}}.$

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Double Integrals by Ugur Efem in Ugur's workspace.

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mass of tetrahedron

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