Problem 65 - Calculus Early Transcendentals

Question

Verify that the volume of a right circular cone with a base radius of \(r\) and a height of \(h\) is \(\pi r^{2} h / 3 .\) Use the region bounded by the line \(y=r x / h,\) the \(x\) -axis, and the line \(x=h,\) where the region is rotated about the \(x\) -axis. Then (a) use the disk method and integrate with respect to \(x,\) and (b) use the shell method and integrate with respect to \(y\)

Step-by-Step Solution

Answer

Based on both the disk method and the shell method, the volume of a right circular cone with a base radius \(r\) and height \(h\) is confirmed to be given by the formula \(\frac{\pi r^2 h}{3}\).

Step 1: 1. Set up the function describing the cone

The cone's base is a circle with radius \(r\) and height \(h\). The equation describing this cone is \(y = \frac{rx}{h}\).

Step 2: 2. Find the radius of each disk

We will divide the cone into thin disks perpendicular to the x-axis. The radius of each disk is determined by the function we found earlier, \(y(x) = \frac{rx}{h}\).

Step 3: 3. Set up the integral with respect to x using the disk method

To find the volume of the cone based on the volume of each disk, we can integrate the disks' areas over the interval \([0, h]\). The volume of each disk is given by the formula \(V(x) = \pi [y(x)]^2 dx\), so our integral would be: $$ V = \int_0^h \pi \left(\frac{rx}{h}\right)^2 dx $$

Step 4: 4. Evaluate the integral

Now, we can evaluate the integral: $$ V = \int_0^h \pi \left(\frac{r^2 x^2}{h^2}\right) dx = \pi \frac{r^2}{h^2} \int_0^h x^2 dx $$ Applying the power rule, we get: $$ V = \pi \frac{r^2}{h^2} \cdot \frac{h^3}{3} = \frac{\pi r^2 h}{3} $$ Therefore, the volume of the cone using the disk method is \(\frac{\pi r^2 h}{3}\). (b) Shell method:

Step 5: 1. Determine the radius and height of each cylindrical shell

We will now divide the cone into thin shells by rotating the region \(x = h - y\frac{h}{r}\). The radius of each shell is \(x\) and the height is \(y\).

Step 6: 2. Set up the integral with respect to y using the shell method

To find the volume of the cone based on the volume of each shell, we can integrate the shells' volumes over the interval \([0, r]\). The volume of each shell is given by the formula \(V(y) = 2\pi x(y) y dy\), so our integral would be: $$ V = \int_0^r 2\pi \left(h - y\frac{h}{r}\right)y dy $$

Step 7: 3. Evaluate the integral

Now, we can evaluate the integral: $$ V = 2\pi h \int_0^r y \left(1 - \frac{y}{r}\right) dy = 2\pi h \int_0^r \left(y - \frac{y^2}{r}\right) dy $$ Separating the terms and applying the power rule, we get: $$ V = 2\pi h \left( \int_0^r y dy - \frac{1}{r} \int_0^r y^2 dy \right) = 2\pi h \left( \frac{r^2}{2} - \frac{r^3}{3r} \right) $$ Simplifying, we obtain: $$ V = 2\pi h \left( \frac{r^2}{2} - \frac{r^2}{3} \right) = 2\pi h \cdot \frac{r^2}{6} = \frac{\pi r^2 h}{3} $$ Therefore, the volume of the cone using the shell method is \(\frac{\pi r^2 h}{3}\).

Key Concepts

Disk Method

When faced with the challenge of calculating the volume of a cone, the disk method provides an effective approach. This technique involves slicing the cone horizontally to form flat, disc-shaped pieces. Each of these disks can be thought of as a very thin cylinder, with a volume equal to the area of the circle \times the thickness of the disk.

During the integration process, the area of each disk is based on the radius, which varies depending on its position along the height, or the x-axis. This radius can be expressed as a function, as seen in the exercise, where the radius is represented by the simple linear equation \(y = \frac{rx}{h}\). Integration is then performed across the range from the cone's tip at x=0 to its base at x=h, culminating in the formula for the volume of the cone. The disk method is a classic example of using integration techniques to compute volumes of solids of revolution in calculus.

Shell Method

The shell method stands as an alternative to the disk method, especially helpful in certain situations where the disk method may be cumbersome. This approach divides the solid into cylindrical shells instead of disks. For a cone, imagine peeling off a thin layer and opening it up to form a cylindrical shell.

In the shell method, one integrates with respect to the y-axis, and the radius of each shell is now the distance from any point on the shell to the y-axis, which is x. The height of the shell corresponds to the value of the function y at that point. This results in an integral that covers the range from y=0 at the base of the cone to y=r at the tip. Upon evaluation, one will again arrive at the volume of the cone. Shifting between disk and shell methods empowers students with flexibility and deepens their grasp of integration techniques and solids of revolution in calculus.

Integration Techniques

Mastering integration techniques is essential for solving complex problems in calculus. These techniques, which include methods like substitution, integration by parts, partial fractions, and the power rule, are instrumental in finding the antiderivative of functions.

Within the context of the exercise, the power rule becomes particularly important. The power rule states that the integral of \(x^n\) with respect to x is \(\frac{x^{n+1}}{n+1}\), provided that n is not equal to -1. This rule simplifies the process of integrating functions like \(x^2\), as seen when determining the volume of the cone using both disk and shell methods. Grasping these techniques is key to success in calculus and is widely applied across different areas in mathematics and physics.

Solid of Revolution

A solid of revolution is a three-dimensional body that's created by rotating a two-dimensional plane area around an axis. The volume of a solid of revolution can be calculated using mathematical methods such as the disk or shell method.

In our exercise, the solid of revolution is a right circular cone. By rotating a triangle around one of its sides (the x-axis in this case), we generate a cone. Visualizing the problem in this way helps in setting up the correct integral for calculating the volume. Understanding the concept of solids of revolution provides valuable insight into how three-dimensional shapes can be formed from simple two-dimensional areas, which is a fascinating application of calculus.

Calculus

Calculus, a branch of mathematics focused on change and motion, is divided into two main areas: differential calculus and integral calculus. Differential calculus deals with the concept of a derivative, which measures how a function changes as its input changes. In contrast, integral calculus focuses on the concept of an integral, providing a way to calculate areas, volumes, and other concepts related to accumulation.

In the exercise, integral calculus is at the forefront, with the integration of a function representing the accumulated sum of areas to find the volume of a solid. Whether using the disk or shell method, calculus enables us to solve complex geometric problems by breaking them down into infinitesimally small pieces, which can then be summed to find a total value, as demonstrated in the solved exercise.