Figure 46

Solution Figure 46(a) shows the solid $E$ that is $xy$-simple. It is enclosed by the surfaces $z=z_{1}(x, y)=0$ and $z=z_{2}(x, y)=1-x-y$, and its projection onto the $xy$-plane is the closed, bounded $x$-simple region $R$ defined by the lines $y=0$ and $y=1-x,$ where $0\leq x\leq 1$. See Figure 46(b).

(a) Since the mass density $\rho =\rho (x, y,z)$ is proportional to the distance from the $yz$-plane, we have $\rho (x, y,z) =kx,$ where $k$ is the constant of proportionality. Then the mass $M$ is $$M=\iiint\limits_{\kern-2ptE}\rho (x, y, z) \,{\it dV}=\iiint\limits_{\kern-2ptE}kx\,{\it dV}=k\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}x \,{\it dz}\,{\it dy}\,{\it dx}$$

Using a CAS, $M=\dfrac{k}{24}.$

(b) The center of mass $(\bar{x}, \bar{y}, \bar{z})$ is $$\begin{eqnarray*} \bar{x}&=&\dfrac{\iiint\limits_{\kern-2ptE}x\rho ( x, y,z) \,{\it dV}}{M}=\dfrac{ \iiint\limits_{\kern-2ptE}xkx\,{\it dV}}{\dfrac{k}{24}}=24\iiint\limits_{\kern-2ptE}x^{2}\,{\it dV}\\[7pt] &=&24\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}x^{2}\,{\it dz}\,{\it dy}\,{\it dx} \end{eqnarray*}$$

Using a CAS, $\overline{x}=\dfrac{2}{5}.$ $$\begin{equation*} \bar{y}=\dfrac{\iiint\limits_{\kern-2ptE}y\rho (x, y,z) \,{\it dV}}{M}=\dfrac{ \iiint\limits_{\kern-2ptE}ykx\,{\it dV}}{\dfrac{k}{24}}=24\int_{0}^{1}\int_{0}^{1-x} \int_{0}^{1-x-y}{\it xy}\,{\it dz}\,{\it dy}\,{\it dx} \end{equation*}$$

Using a CAS, $\bar{y}=$ $\dfrac{1}{5}.$

946

$$\begin{equation*} \bar{z}=\dfrac{\iiint\limits_{\kern-2ptE}z\rho (x, y,z) \,{\it dV}}{M}=\dfrac{ \iiint\limits_{\kern-2ptE}zkx\,{\it dV}}{\dfrac{k}{24}}=24\int_{0}^{1}\int_{0}^{1-x} \int_{0}^{1-x-y}xz\,{\it dz}\,{\it dy}\,{\it dx} \end{equation*}$$

Using a CAS $\ \bar{z}=$ $\dfrac{1}{5}.$

The center of mass of the tetrahedron is located at the point $\left( \dfrac{2}{5},\dfrac{1}{5},\dfrac{1}{5}\right).$