Problem 29 - Calculus Early Transcendentals

Question

Find the center of mass of the following solids, assuming a constant density of 1. Sketch the region and indicate the location of the centroid. Use symmetry when possible and choose a convenient coordinate system. The tetrahedron in the first octant bounded by \(z=1-x-y\) and the coordinate planes

Step-by-Step Solution

Answer

Answer: The coordinates of the centroid of the tetrahedron are (\(\frac{1}{6}, \frac{1}{6}, \frac{1}{6}\)).

Step 1: Sketch the region and identify symmetries

Sketch the given tetrahedron in the first octant of the xyz-coordinate system. It is represented by the vertices (1,0,0), (0,1,0), (0,0,1), and (0,0,0). From the sketch, we observe that the tetrahedron is symmetric about the plane \(x=y\), \(x=z\), and \(y=z\). We can use these symmetries to simplify calculations of the centroid coordinates.

Step 2: Set up integrals for centroid coordinates

To find the centroid coordinates, we have to compute the volume V and the integral of each coordinate with respect to the region. Due to the tetrahedron's symmetry, we can obtain the centroid coordinates (\(\bar{x},\bar{y},\bar{z}\)) directly, as they will all be the same. The centroid coordinates are given by: \(\bar{x} = \bar{y} = \bar{z} = \frac{1}{V}\int\int\int_R x\:dV\)

Step 3: Compute the volume of the tetrahedron

The volume of the tetrahedron can be calculated using triple integration and the region R: \(V =\int\int\int_R dV = \int_0^1\int_0^{1-x}\int_0^{1-x-y}dzdydx\) To calculate the volume, evaluate the integral step by step: \(V = \int_0^1\int_0^{1-x}(1-x-y)dydx = \int_0^1\left[-\frac{1}{2}y^2+((1-x)y-\frac{1}{2}y^2)\right]_{y=0}^{y=1-x}dx\) \(V = \int_0^1\frac{1}{2}x^2dx = \frac{1}{6}\) Thus, the volume of the tetrahedron is \(\frac{1}{6}\).

Step 4: Compute the integral of the x-coordinate

Now we need to compute the integral of x with respect to the region R: \(\int\int\int_R x\:dV = \int_0^1\int_0^{1-x}\int_0^{1-x-y} x\:dzdydx\) Evaluate the integral like before: \(\int\int\int_R x\:dV = \int_0^1\int_0^{1-x}(x(1-x-y))dydx = \int_0^1x(1-x)^2dx\) After evaluating the integral, we have: \(\int\int\int_R x\:dV = \frac{1}{24}\)

Step 5: Compute the centroid coordinates

Finally, calculate the coordinates of the centroid using the computed values: \(\bar{x} = \bar{y} = \bar{z} = \frac{1}{V}\int\int\int_R x\:dV = 4\cdot\frac{1}{24} = \frac{1}{6}\) Hence, the centroid of the tetrahedron is located at (\(\frac{1}{6}, \frac{1}{6}, \frac{1}{6}\)).