Problem 45 - Thomas Calculus

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 \(\begin{array}{ll}{\text { a. the } x \text { -axis. }} & {\text { b. the } y \text { -axis. }} \\ {\text { c. the line } y=2 .} & {\text { d. the line } x=4}\end{array}\)

Step-by-Step Solution

Answer

Volumes: (a) \(8\pi\); (b) \(8\pi\); (c) \(\frac{16}{3}\pi\); (d) \(\frac{128}{15}\pi\).

Step 1: Find the Bounds for the Volume

To solve for the volume of the revolution around any line, first determine the bounds for the integration. For the function \( y = \sqrt{x} \), the line \( y = 2 \) corresponds to \( x = 4 \), because \( y = \sqrt{x} \Rightarrow y^2 = x \), so at \( y = 2 \), \( x = 4 \). Therefore, the region is bounded between \( x = 0 \) and \( x = 4 \).

Step 2: Volume about the x-axis

When revolving around the \( x \)-axis, use the disk method. The radius of the disk at any x-value is equal to \( y \) which is \( \sqrt{x} \). The volume is given by the integral:\[ V = \pi \int_{0}^{4} (\sqrt{x})^2 \,dx = \pi \int_{0}^{4} x \,dx \]Integrate to get: \[ \pi \left[ \frac{x^2}{2} \right]_{0}^{4} = \pi \left( \frac{16}{2} \right) = 8\pi \]

Step 3: Volume about the y-axis

Use the method of cylindrical shells. The height of a shell at each \( y \) is \( 4 - y^2 \): \[ V = 2\pi \int_{0}^{2} y(4 - y^2) \,dy \]Integrate to find: \[ = 2\pi \left[ 4\frac{y^2}{2} - \frac{y^4}{4} \right]_{0}^{2} = 2\pi \left( 8 - 4 \right) = 8\pi \]

Step 4: Volume about the line y=2

Using the washer method, the outer radius \( R \) is 2 and the inner radius \( r \) is \( 2 - \sqrt{x} \): \[ V = \pi \int_{0}^{4} ((2)^2 - (2-\sqrt{x})^2) \,dx \]Simplify the integrand and solve:\[ = \pi \int_{0}^{4} (4 - 4 + 4\sqrt{x} - x) \,dx = \pi \int_{0}^{4} (4\sqrt{x} - x) \,dx \]\[ = \pi \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{2} \right]_{0}^{4} = \pi \left( \frac{64}{3} - 8 \right) = \frac{16}{3}\pi \]

Step 5: Volume about the line x=4

Utilize the shell method for revolving around \( x = 4 \). The radius of each shell is \( 4 - x \), and the height is \( \sqrt{x} \): \[ V = 2\pi \int_{0}^{4} (4-x)\sqrt{x} \,dx \]Integrate:\[ = 2\pi \int_{0}^{4} (4x^{1/2} - x^{3/2}) \,dx \]\[ = 2\pi \left[ \frac{8}{3}x^{3/2} - \frac{2}{5}x^{5/2} \right]_{0}^{4} = 2\pi \left( \frac{64}{3} - \frac{64}{5} \right) = \frac{128}{15}\pi \]

Key Concepts

Volume by Integration

Calculating the volume of solids of revolution is an intriguing application of integral calculus. This technique involves revolving a two-dimensional region around an axis, generating a three-dimensional solid object. To find the volume of such a solid, integration is used. This advanced mathematical concept allows us to sum an infinite number of infinitesimally thin slices of the solid. Each slice, taken perpendicular to the axis of rotation, contributes to the total volume. The basic idea is to compute the integral of a function that represents the cross-sectional area of these slices over a specified range. The result is the total volume. Different methods of integration are employed based on the axis of rotation, including the disk, washer, and shell methods.

Disk Method

The disk method is a straightforward technique to find volumes of revolution, particularly useful when a region is revolved around a horizontal or vertical axis, like the x-axis.For the disk method:

• Visualize the solid as a stack of small disks.
• Each disk has a small thickness (denoted by \( dx \) or \( dy \)) and a radius determined by the function \( y = f(x) \) or its counterpart.

The volume of each disk is essentially the area of the circular face, \( \pi r^2 \), times the thickness.The total volume is obtained by integrating these small volumes across the bounds of the region:\[ V = \pi \int_a^b [f(x)]^2 \, dx \]Where \( a \) and \( b \) denote integration limits.This method was employed when finding the volume of the region bounded by \( y = \sqrt{x} \) rotated around the x-axis, giving the volume as \( 8\pi \).

Shell Method

The shell method is effective when revolving around a vertical line, such as the y-axis. It's particularly advantageous for functions easier expressed in terms of y, avoiding solving for x explicitly.This method involves:

• Imagining the solid as made of concentric cylindrical shells.
• Each shell is defined by a height \( h \) and radius \( r \), varying with the independent variable.

The volume of one shell is calculated by wrapping the height around the circumference of the shell, then multiplying by the thickness.The formula for the shell method is:\[ V = 2\pi \int_c^d r(y)h(y) \, dy \]Here, the shell method helped calculate volumes around both the y-axis and the line \( x = 4 \), yielding results of \( 8\pi \) and \( \frac{128}{15}\pi \) respectively.

Washer Method

The washer method is a variation of the disk method, applied when the region rotated results in a hollow core, such as revolving around non-axis lines like \( y = 2 \).It resembles the disk method but accounts for an inner radius, leaving an empty space. Each slice resembles a washer:

• The volume of a washer is the area of the outer disk minus the area of the inner disk.
• Integration of these individual volumes gives the total volume.

The washer method uses:\[ V = \pi \int_a^b \, (R^2 - r^2) \, dx \]Where \( R \) is the outer radius and \( r \) is the inner radius.When revolving the region around \( y = 2 \), this method yields a volume of \( \frac{16}{3}\pi \), demonstrating its effectiveness with hollow solids.