Question

Find the mass of the solid in the shape of the tetrahedron that is bounded by the planes x + y + z = 1 , xy- plane, xz- plane and yz- plane if its density is given by 𝛿(x, y, z) = x 

Answer 1

The mass of the solid region $E$ is evaluated as,

$m=\iiint_{E}\delta(x,y,z)dV$ ,

where $E=\lbrace(x,y,z)|0\eqslantless x \eqslantless 1, 0 \eqslantless y \eqslantless 1-x, 0 \eqslantless z \eqslantless1-x-y\rbrace$

and $\delta(x,y,z)=x$

So, the mass of the solid is,

$m=\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}(x)dzdydx$

$=\int_{0}^{1}\int_{0}^{1-x}x(1-x-y)dydx$

$=\int_{0}^{1}x[(1-x)y-\frac{y^2}{2}]_{y=0}^{1-x}dx$

$=\int_{0}^{1}x[(1-x)(1-x)-\frac{(1-x)^2}{2}]dx$

$=\int_{0}^{1}x[(1-x)^2-\frac{(1-x)^2}{2}]dx$

$=\frac{1}{2}\int_{0}^{1}x(1-x)^2dx$

$=\frac{1}{2}\int_{0}^{1}x(1+x^2-2x)dx$

$=\frac{1}{2}\int_{0}^{1}(x+x^3-2x^2)dx$

$=\frac{1}{2}[\frac{x^2}{2}+\frac{x^4}{4}-2(\frac{x^3}{3})]_{0}^{1}$

$=\frac{1}{2}[\frac{1}{2}+\frac{1}{4}-\frac{2}{3}]$

$=\frac{1}{2}(\frac{1}{12})$

$=\frac{1}{24}$

Therefore, the mass of the solid region $E$ is $m=\iiint_{E}\delta(x,y,z)dV=\frac{1}{24}$