Problem 47 - Thomas Calculus in SI Units

Question

Find the volume of the solid generated by revolving the region bounded by \(y=\sqrt{x}\) and the lines \(y=2\) and \(x=0\) about a. the \(x\) -axis. b. the \(y\) -axis. c. the line \(y=2\) d. the line \(x=4\)

Step-by-Step Solution

Answer

a. \(8\pi\); b. \(4\pi\); c. \(\frac{16\pi}{3}\); d. \(8\pi\).

Step 1: Identify the Functions and Boundaries

The region of interest is bounded by the curve \( y = \sqrt{x} \), the line \( y = 2 \), and the line \( x = 0 \). This means we are dealing with a region from \( x = 0 \) to \( x = 4 \) (since \( \sqrt{x} = 2 \) implies \( x = 4 \)).

Step 2a: Volume Around the x-axis - Disk Method

The volume when revolving around the \( x \)-axis can be calculated using the Disk Method. The formula is \( V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx \). Here, \((\sqrt{x})^2 = x\). So, \( V = \pi \int_{0}^{4} x \, dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \times \frac{16}{2} = 8\pi \).

Step 2b: Volume Around the y-axis - Shell Method

Use the Shell Method for rotation around the \( y \)-axis. The formula is \( V = 2\pi \int_{0}^{2} y (y^2) \, dy \). Solving, \( y^2 = x \) gives \( V = 2\pi \int_{0}^{2} y (x) \, dy \), with \( x = y^2 \), leading to \( V = 2\pi \int_{0}^{2} y(y^2) \, dy = 2\pi \int_{0}^{2} y^3 \, dy = 2\pi \left[ \frac{y^4}{4} \right]_0^2 = 4\pi \).

Step 2c: Volume Around the Line y=2 - Washer Method

When revolving around the line \( y=2 \), we use the Washer Method. Consider the outer radius \( R(y) = 2 \) and inner radius \( r(y) = y \). The volume formula is \( V = \pi \int_{0}^{2} (2^2 - y^2) \, dy = \pi \left[ 4y - \frac{y^3}{3} \right]_0^2 = \pi (8 - \frac{8}{3}) = \frac{16\pi}{3} \).

Step 2d: Volume Around the Line x=4 - Shell Method

For the line \( x=4 \), use the Shell Method again. The formula is \( V = 2\pi \int_{0}^{2} (4-y^2) y \, dy \). Simplify and integrate: \( V = 2\pi \left[ 4\frac{y^2}{2} - \frac{y^4}{4} \right]_0^2 = 2\pi [8 - 4] = 8\pi \).

Key Concepts

Disk Method

The Disk Method is one of the core techniques in Calculus for finding the volume of a solid of revolution. This method is particularly useful when the solid is created by rotating a region around an axis that directly involves the function itself. In essence, the Disk Method involves slicing the solid into thin, circular disks, and then calculating the volume of each slice before summing them up. The formula to find the volume using the Disk Method is given by:
\[ V = \pi \int_a^b [f(x)]^2 \ dx \]
Here, \( f(x) \) represents the function that defines the outer edge of the solid, and \( a \) and \( b \) are the limits of integration, corresponding to the interval along the axis of rotation.
Using this method, for the given problem, when revolving the region about the \( x \)-axis, the resulting volume is:

• Identify the function: \( y = \sqrt{x} \), which means \( (\sqrt{x})^2 = x \).
• Calculate the definite integral: \[ V = \pi \int_0^4 x \ dx = 8\pi \]

This calculation demonstrates how integrals can effectively represent the accumulation of infinitely many disks to create a solid.

Shell Method

The Shell Method is an alternative technique to the Disk Method, especially useful when revolving around an axis parallel to the axis of the function. In this approach, the solid is divided into cylindrical shells rather than disks. Each shell's volume can be calculated and summed up using integration. The formula for volume using the Shell Method is:
\[ V = 2\pi \int_a^b x(f(x)) \, dx \]
Here, \( x \) acts as the radius of each shell, while \( f(x) \) represents the height.
For the problem of revolving around the \( y \)-axis:

• Recognize that the function can be expressed as \( x = y^2 \).
• The formula for shells becomes \[ V = 2\pi \int_0^2 y(y^2) \, dy = 4\pi \]

Operating with shells, you effectively wrap the area around the axis, resulting in a cylindrical formation. This method shows how changing the axis of rotation requires a different approach but still uses the fundamentals of integration.

Washer Method

The Washer Method expands on the Disk Method by accommodating a hollow space in the middle of the solid, much like a washer. This method is typically applied when there's a rotation around a line that creates both an inner and outer boundary. The volume here is calculated by taking the difference between the two radii disks; hence, the name "Washer Method."
The formula for evaluating the volume is:
\[ V = \pi \int_a^b ([R(y)]^2 - [r(y)]^2) \ dy \]
Where \( R(y) \) is the outer radius and \( r(y) \) is the inner radius.
In this exercise, when rotating around the line \( y = 2 \):

• Identify the outer radius: \( R(y) = 2 \).
• Identify the inner radius: \( r(y) = y \).
• Compute the integral: \[ V = \pi \int_0^2 (2^2 - y^2) \, dy = \frac{16\pi}{3} \]

This approach helps visualize the process of volume subtraction, showing how integration can effectively carve out spaces to yield accurate results.

Integral Calculus

Integral Calculus is at the heart of calculating volumes of revolution. It focuses on the area under curves and the accumulation of quantities, which are essential when dealing with rotations and volumes. The two primary operations in Integral Calculus are definite and indefinite integration.
In our context, definite integrals help evaluate the accumulated volume generated by rotating curves and lines around a certain axis. By setting up appropriate limits and working through the integral, Calculus allows us to find exact volumes, as demonstrated in this exercise.
Some critical points in using Integral Calculus for volumes include:

• Determining the function that accurately represents the boundary line of the solid.
• Understanding the axis of rotation to choose suitable methods, such as the Disk, Shell, or Washer Method.
• Setting proper integration limits based on the problem's geometry.

Utilizing these techniques, Integral Calculus provides a powerful toolset to tackle a wide array of volume-related problems, enabling learners to connect geometric intuition with mathematical precision.