Problem 55 - 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 cone has a base with a radius of \(a\) and a height of \(h\). How far from the base is the center of mass?

Step-by-Step Solution

Answer

Answer: The center of mass of a solid cone with base radius \(a\) and height \(h\) is at a distance of \(\frac{h}{4}\) from the base.

Step 1: Choosing a coordinate system and identifying the boundaries

We can choose a cylindrical coordinate system \((r, \theta, z)\) for this problem, which is convenient for a cone. In this system, the base of the cone lies in the \(z = 0\) plane, and its apex is at \(z = h\). The lateral surface of the cone is given by \(r = \frac{a}{h}z\).

Step 2: Computing the volume of the cone

Before we compute the center of mass, we need to find the volume of the cone. The volume of a cone is given by the formula \(V = \frac{1}{3} \pi a^2 h\).

Step 3: Defining the differential volume element

In the cylindrical coordinate system, a differential volume element is given by \(dV = r dr d\theta dz\). We will use this to compute the center of mass.

Step 4: Computing the center of mass

The center of mass is given by the following integral: $$z_{cm} = \frac{1}{V} \int_0^h \int_0^{\frac{a}{h}z} \int_0^{2\pi} z r dr d\theta dz$$

Step 5: Evaluating the integral

We can evaluate the integral step by step: 1. Integrate with respect to \(r\) $$\int_0^{\frac{a}{h}z} z r dr = \frac{1}{2}z^3\frac{a^2}{h^2}$$ 2. Integrate with respect to \(\theta\) $$\int_0^{2\pi}\frac{1}{2}z^3\frac{a^2}{h^2} d\theta = \frac{1}{2}z^3\frac{a^2}{h^2} \cdot 2\pi = \pi z^3\frac{a^2}{h^2}$$ 3. Integrate with respect to \(z\) $$\int_0^h \pi z^3\frac{a^2}{h^2} dz = \pi \frac{a^2}{h^2} \cdot \frac{1}{4} h^4 = \frac{1}{4}\pi a^2 h^2$$

Step 6: Computing the distance from the base

Finally, we can compute the distance from the base by dividing by the volume: $$z_{cm} = \frac{1}{V} \frac{1}{4}\pi a^2 h^2$$ $$z_{cm} = \frac{\frac{1}{4}\pi a^2 h^2}{\frac{1}{3} \pi a^2 h}$$ $$z_{cm} = \frac{h}{4}$$ So, the center of mass of the cone is \(\frac{h}{4}\) from the base.