Problem 54 - Calculus Early Transcendentals

Question

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A solid rectangular box has sides of length \(a, b,\) and \(c .\) Where is the center of mass relative to the faces of the box?

Step-by-Step Solution

Answer

Answer: The coordinates of the center of mass for the solid rectangular box are \(\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right)\).

Step 1: Identify the bounding surfaces and set a coordinate system

The solid rectangular box is defined by six surfaces: the bottom, top, left, right, front, and back faces of the box. We will use a Cartesian coordinate system with x, y, and z axes. We can define the box such that its corners are located at \((0, 0, 0),\) \((0, b, 0),\) \((a, 0, 0),\) \((0, 0, c),\) \((a, b, 0),\) \((0, b, c),\) \((a, 0, c),\) and \((a, b, c).\)

Step 2: Determine the volume of the box

Since the box is a rectangular box, the volume can be found by multiplying the length, width, and height. In this case, the volume V is given by: $$ V = a \cdot b \cdot c$$

Step 3: Compute the center of mass

To find the center of mass for the solid rectangular box, we will calculate the center in each axis (x, y, and z). The center of mass coordinates is given by \(\bar{x}, \bar{y},\) and \(\bar{z}.\) Since the density is constant, we can use the average of the coordinates of the vertices for each axis. For the x-axis: $$\bar{x} = \frac{0 + 0 + a + 0 + a + 0 + a + a}{8} = \frac{4a}{8} = \frac{a}{2}$$ For the y-axis: $$\bar{y} = \frac{0 + b + 0 + 0 + b + b + 0 + b}{8} = \frac{4b}{8} = \frac{b}{2}$$ For the z-axis: $$\bar{z} = \frac{0 + 0 + 0 + c + 0 + c + c + c}{8} = \frac{4c}{8} = \frac{c}{2}$$

Step 4: Present the center of mass

Now that we have found the center of mass for each axis, we can express the coordinates of the center of mass relative to the faces of the box as: Center of mass: \(\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right)\)