91Ó°ÊÓ

The tetrahedron is bounded by the following planes: 1. \(x=0\) (yz-plane) 2. \(y=0\) (xz-plane) 3. \(z=0\) (xy-plane) 4. \(x / a+y / a+z / a=1\)

To find the volume of the tetrahedron, we can integrate over the region using triple integration. The limits of integration for x, y, and z can be found by analyzing the bounding surfaces. We have: 1. \(x\) ranges from 0 to \(a\). 2. For each fixed \(x\), \(y\) ranges from 0 to \(a(1 - x/a)\). 3. For each fixed \(x\) and \(y\), \(z\) ranges from 0 to \(a(1 - x/a - y/a)\). Now we can evaluate the triple integral to find the volume: \(\text{Volume} = \int_{0}^{a} \int_{0}^{a(1-\frac{x}{a})} \int_{0}^{a(1 - \frac{x}{a} - \frac{y}{a})} dz dy dx\) Solving the integral gives us \(\text{Volume}=\frac{1}{6}a^3\).

Now, we will find the \(x\), \(y\), and \(z\) coordinates of the center of mass of the tetrahedron, \((\bar{x}, \bar{y}, \bar{z})\). We need to evaluate the triple integrals of \(x\), \(y\), and \(z\) over the tetrahedron and then divide by the volume found in Step 2. I. \(\bar{x} = \frac{1}{\text{Volume}}\int_{0}^{a} \int_{0}^{a(1-\frac{x}{a})} \int_{0}^{a(1 - \frac{x}{a} - \frac{y}{a})} x dz dy dx\) II. \(\bar{y} = \frac{1}{\text{Volume}}\int_{0}^{a} \int_{0}^{a(1-\frac{x}{a})} \int_{0}^{a(1 - \frac{x}{a} - \frac{y}{a})} y dz dy dx\) III. \(\bar{z} = \frac{1}{\text{Volume}}\int_{0}^{a} \int_{0}^{a(1-\frac{x}{a})} \int_{0}^{a(1 - \frac{x}{a} - \frac{y}{a})} z dz dy dx\) Evaluating these integrals gives: I. \(\bar{x} = \frac{1}{4}a\) II. \(\bar{y} = \frac{1}{4}a\) III. \(\bar{z} = \frac{1}{4}a\) Finally, the coordinates of the center of mass are \((\frac{1}{4}a, \frac{1}{4}a, \frac{1}{4}a)\).