Problem 32 - Thomas Calculus in SI Units

Question

Find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so. The region bounded by \(y=\sqrt{x}, y=2, x=0\) about a. the \(x\) -axis b. the \(y\) -axis c. the line \(x=4\) d. the line \(y=2\)

Step-by-Step Solution

Answer

a. \(8\pi\), b. \(\frac{8\pi}{3}\), c. \(\frac{40\pi}{3}\), d. \(\frac{8\pi}{3}\)

Step 1: Understanding the problem

We are asked to find the volume of solids of revolution. Each part involves rotating a region around a certain axis. The region of interest is bounded by the curves \( y = \sqrt{x} \), \( y = 2 \), and \( x = 0 \). We will use the disk or washer method depending on the axis of rotation.

Step 2: Part (a): Revolving around the x-axis

The solid is formed by revolving the region between \( y=\sqrt{x} \) and \( y=2 \) around the x-axis. The volume \( V \) is given by the integral \( V = \int_{0}^{4} \pi (R^2 - r^2) \, dx \) where the outer radius \( R = 2 \) and the inner radius \( r = \sqrt{x} \). Hence, the integral becomes \( V = \int_{0}^{4} \pi (4 - x) \, dx \).

Step 3: Calculate Part (a): Integral

Solve the integral: \( V = \pi \int_{0}^{4} (4 - x) \, dx = \pi [4x - \frac{x^2}{2}] \Big|_0^4 = \pi [16 - 8] = 8\pi \). Thus, the volume is \( 8\pi \).

Step 4: Part (b): Revolving around the y-axis

Revolve around the y-axis using the shell method. The volume is \( V = \int_{0}^{2} 2\pi x \cdot (2 - \sqrt{x}) \, dx \). To solve this, express \( x \) in terms of \( y \) as \( x = y^2 \). Consequently, the integral becomes \( V = \int_{0}^{2} 2\pi y^2 (2 - y) \, dy \).

Step 5: Calculate Part (b): Integral

Calculate the integral \( V = 2\pi \int_{0}^{2} (2y^2 - y^3) \, dy = 2\pi \left[\frac{2y^3}{3} - \frac{y^4}{4}\right]_0^{2} = 2\pi [\frac{16}{3} - 4] = \frac{8\pi}{3} \). Thus, the volume is \( \frac{8\pi}{3} \).

Step 6: Part (c): Revolving around x = 4

The solid is formed by revolving around \( x = 4 \). The volume \( V \) is \( \int_{0}^{4} \pi ((4 - \sqrt{x})^2 - 0^2) \, dx \). The radius is \( 4 - x \).

Step 7: Calculate Part (c): Integral

The integral becomes \( V = \pi \int_{0}^{4} (4 - \sqrt{x})^2 \, dx \). This evaluates to \( \pi \left[ 16x - \frac{8x^{3/2}}{3} + \frac{x^2}{2} \right]_0^4 = \pi [64 - \frac{128}{3} + 8] = \frac{40\pi}{3} \). Thus, the volume is \( \frac{40\pi}{3} \).

Step 8: Part (d): Revolving around y = 2

For rotation around \( y = 2 \), the washer method is ideal. The volume \( V \) is \( \int_{0}^{4} \pi ( (2 - \sqrt{x})^2 - 0^2 ) \, dx \). Radii are \( 2 - x \).

Step 9: Calculate Part (d): Integral

Calculate the integral \( V = \pi \int_{0}^{4} (2 - \sqrt{x})^2 \, dx \). This evaluates to \( \pi \left[ 4x - \frac{4x^{3/2}}{3} + \frac{x^2}{4} \right]_0^4 = \pi [16 - \frac{32}{3} + 4] = \frac{8\pi}{3} \). Thus, the volume is \( \frac{8\pi}{3} \).

Key Concepts

Disk Method

The Disk Method is a way to calculate the volume of a solid of revolution, which is formed by rotating a region around an axis. Essentially, it involves slicing the solid perpendicular to the axis of rotation into thin disks. By summing up the volumes of these circular disks, you get the volume of the entire solid.

Here's how it works in detail:

• Cross section: Each disk's cross-section is a circle with a radius corresponding to the value of the function at a particular point.
• Volume of a disk: The volume of a single disk with radius R and thickness dx is given by the formula: \[ V_{disk} = \pi R^2 \, dx \]
• Integrating the disks: Integrate the disk volumes across the interval. The definite integral, \[ V = \pi \int_{a}^{b} R(x)^2 \, dx \], gives the total volume.

We used this method to calculate the volume when we revolved a region around the x-axis. The integration process involved finding the outer radius since there was no inner radius here.

Washer Method

The Washer Method is an extension of the Disk Method suitable when there is a hollow central region. Just think of it as a disk with a hole in the middle, resembling a washer. This method is used when the region enclosed goes from an outer curve to an inner curve, making it perfect for finding volumes of solids that have gaps.

Key steps include:

• Cross-section setup: Define both the outer radius R and the inner radius r for the washers formed.
• Volume of a washer: Subtract the volume of the inner disk from the volume of the outer disk: \[ V_{washer} = \pi (R^2 - r^2) \, dx \]
• Integral for volume: Integrate the washer volume from two bounding curves: \[ V = \pi \int_{a}^{b} (R(x)^2 - r(x)^2) \, dx \].

In our exercise, this method was used to calculate volumes revolving around axes where different bounding curves form the solid. By setting up the outer and inner functions, you achieve precision for hollow solids.

Shell Method

The Shell Method is another powerful technique for volumes of revolution, especially useful when dealing with vertical strips. It works by considering cylindrical shells formed by revolving a region along an axis. If you find the Disk Method a bit cumbersome in certain situations, the Shell Method provides a handy alternative.

Concepts you need to understand:

• Shell setup: Use vertical or horizontal lines to set up a cylindrical shell.
• Volume of a shell: The volume for each shell is given by: \[ V_{shell} = 2\pi (radius)(height)(thickness) \].
• Cylindrical shell equation: The integral is formulated as: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dx \].

When no natural disk or washer fits easily, shells can simplify the process. In our problem, the Shell Method was used to revolve around the y-axis. The method allowed for straightforward integration by utilizing appropriate functions.

Definite Integrals

Definite Integrals are at the heart of calculating volumes in calculus, especially with solids of revolution. They allow us to sum an infinite number of infinitesimally small elements, like disks or shells, over a specified interval. This calculation leads to precise volume measurements.

• Understanding definite integrals: It provides accumulation of quantities, such as area, volume, or other physical properties over an interval.
• Borders of integration: Defined by the interval \[ a, b \]. These bounds specify the start and end values of the rotation.
  
• Application in volumes: By integrating the cross-sectional areas, you find the total volume of the solid of revolution. Use limits to sum continuous cross-sections accurately.

All methods discussed utilize definite integrals extensively. Whether in the Disk, Washer, or Shell Method, they convert local geometry into whole-volume results, showcasing calculus's power.