# MvCalc57

Mass of the tetrahedron=$\iiint_V \rho dx dy dz\,$ where rho is the constant density of the substance.

$M=\rho\int_0^a\int_{y=0}^{b(1-\frac{x}{a})}\int_{z=0}^{c(1-\frac{x}{a}-\frac{y}{b})}dz dy dx\,$

=$\rho\int_0^a\int_{y=0}^{b(1-\frac{x}{a})} c(1-\frac{x}{a}-\frac{y}{b})dy dx\,$

=$\rho c\int_0^a [(1-\frac{x}{a})y-\frac{y^2}{2b}]_{0}^{b(1-\frac{x}{a})} dx\,$

=$\frac{cb\rho}{2}\int_0^a(1-\frac{x}{a})^2dx=\frac{\rho bc}{2}\frac{a}{3}=\frac{\rho abc}{6}\,$

Let x1,y1,z1 be the coordinates of the centroid.Then

$M x_1=\rho\int_0^a\int_{y=0}^{b(1-\frac{x}{a})}\int_{z=0}^{c(1-\frac{x}{a}-\frac{y}{b})} x dz dy dx\,$

=$\rho\int_0^a\int_{y=0}^{b(1-\frac{x}{a})}cx(1-\frac{x}{a}-\frac{y}{b})dy dx\,$

=$c\rho\int_0^a[x(1-\frac{x}{a})y-\frac{xy^2}{ab}]_{y=0}^{b(1-\frac{x}{a})} dx\,$

=$c\rho b\int_0^a x(1-\frac{x}{a})^2 dx=\rho bc \frac{a^2}{12}\,$

$x_1=\frac{\rho a^2bc}{12}\frac{6}{\rho abc}=\frac{a}{4}\,$

Similarly,$y_1=\frac{b}{4},z_1=\frac{c}{4}\,$

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