Problem 32 - Thomas Calculus

Question

In Exercises \(31-36,\) 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 $$ \begin{array}{ll}{\text { a. the } x \text { -axis }} & {\text { b. the } y \text { -axis }} \\ {\text { c. the line } x=4} & {\text { d. the line } y=2}\end{array} $$

Step-by-Step Solution

Answer

a) \(8\pi\), b) \(\frac{24\pi}{5}\), c) \(\frac{32\pi}{15}\), d) \(\frac{64\pi}{5}\).

Step 1: Understand the Problem

We need to find the volume of solids generated by revolving the region bounded by the curves \( y = \sqrt{x} \), \( y = 2 \), and \( x = 0 \) around different axes and lines. The solid can be found using methods like the disk method or the washer method.

Step 2: Identify Region for X-axis (Part a)

The curves \( y = \sqrt{x} \) and \( y = 2 \) intersect when \( \sqrt{x} = 2 \), which gives \( x = 4 \). This means the region in question for revolution around the x-axis is bounded by \( y = \sqrt{x} \), \( y = 2 \), and \( x = 0 \), from \( x = 0 \) to \( x = 4 \).

Step 3: Solve for X-axis (Part a) Using Disk Method

The volume for part (a) can be calculated using the disk method as follows:\[V = \pi \int_{0}^{4} (2^2 - (\sqrt{x})^2) \; dx = \pi \int_{0}^{4} (4 - x) \; dx\]Calculating this integral gives:\[V = \pi \left[ 4x - \frac{x^2}{2} \right]_{0}^{4} = \pi \left[ (16 - 8) - (0 - 0) \right] = 8\pi\]

Step 4: Identify Region for Y-axis (Part b)

For revolution around the y-axis, the region is bounded by \( y = \sqrt{x} \), \( y = 2 \), and \( x = 0 \). We convert \( y = \sqrt{x} \) to \( x = y^2 \). The limits for \( y \) are from \( 0 \) to \( 2 \).

Step 5: Solve for Y-axis (Part b) Using Washer Method

The volume for part (b) is found with the washer method as:\[V = \pi \int_{0}^{2} ((2)^2 - (y^2)^2) \; dy = \pi \int_{0}^{2} (4 - y^4) \; dy\]Calculating this integral gives:\[V = \pi \left[ 4y - \frac{y^5}{5} \right]_{0}^{2} = \pi \left[ 8 - \frac{32}{5} \right] = \frac{24\pi}{5}\]

Step 6: Identify Region for Line x=4 (Part c)

For revolution around the line \( x = 4 \), the same curves are considered but we revolve about this vertical line. The outer radius is \( 4 - x \) and the inner radius is \( 4 \) minus \( y^2 \).

Step 7: Solve for Line x=4 (Part c) Using Washer Method

The volume for part (c) is calculated using:\[V = \pi \int_{0}^{2} ((4)^2 - (4 - y^2)^2) \; dy\]This becomes:\[V = \pi \int_{0}^{2} (16 - (16 - 8y^2 + y^4)) \; dy = \pi \int_{0}^{2} (8y^2 - y^4) \; dy\]Solving this integral gives:\[V = \pi \left[ \frac{8y^3}{3} - \frac{y^5}{5} \right]_{0}^{2} = \pi \left[ \frac{64}{3} - \frac{32}{5} \right]\]Simplifying, we find:\[V = \frac{32\pi}{15}\]

Step 8: Identify Region for Line y=2 (Part d)

For revolving around \( y = 2 \), the cylindrical shell method is more efficient. Each shell has height \( \sqrt{x} - 0 \), with radius \( 2 - y = 2 - \sqrt{x} \) once replaced with \( x \).

Step 9: Solve for Line y=2 (Part d) Using Shell Method

Volume using cylindrical shells:\[V = 2\pi \int_{0}^{4} (x)(\sqrt{x}) \; dx\]This simplifies to:\[V = 2\pi \int_{0}^{4} x^{3/2} \; dx\]Integrating gives:\[V = 2\pi \left[ \frac{2}{5} x^{5/2} \right]_{0}^{4} = \frac{64\pi}{5}\]

Step 10: Conclusion

The volumes for each of these revolves are calculated by finding limits and using appropriate integration methods specific to the axis of rotation in parts (a) through (d).

Key Concepts

Disk Method

The Disk Method is a powerful technique for finding the volume of a solid of revolution. This method involves rotating a region about an axis to form a solid and slicing the solid into thin disks perpendicular to the axis of revolution. Each disk has a small thickness, represented as \( \Delta x \) or \( \Delta y \), and a radius that depends on the function describing the region being rotated.
To calculate the volume of the solid, integrate the area of these disks from the start to the end of the interval. The formula for the disk method, when rotating around the x-axis, is:
\[V = \pi \int_{a}^{b} [f(x)]^2 \, dx\]For the given exercise, when the region bounded by \( y = \sqrt{x} \), \( y = 2 \), and \( x = 0 \) is revolved around the x-axis, the radius of each disk corresponds to the distance from \( y = \sqrt{x} \) to \( y = 2 \). Therefore, the necessary integral was set up to find the volume from \( x = 0 \) to \( x = 4 \).

Washer Method

The Washer Method is an extension of the disk method, useful when the region to be revolved creates a solid with a hole. Just like a washer has a hole in the middle, solids generated using this method involve subtracting the volume of a smaller disk from a larger one, creating a "washer" shape.
To use this method, you find the volume by subtracting the inner radius from the outer radius squared, integrating over the specified interval:
\[V = \pi \int_{a}^{b} \, [(R_{ ext{outer}})^2 - (R_{ ext{inner}})^2] \, dy\]In this exercise, the Washer Method was applied for the region revolving around the y-axis. The transformation \( x = y^2 \) was used, and integration was carried out over \( y \) from 0 to 2. This method showed its efficiency in handling solids of revolution where an inner radius is subtracted.

Cylindrical Shell Method

The Cylindrical Shell Method is particularly useful when revolving around a line parallel but not touching the region, such as the y-axis. Instead of slicing the solid into disks, imagine wrapping the region into hollow cylindrical shells.
The volume of each shell is determined by the height of the region, the distance from the axis of rotation (radius), and the thickness of the shell. The general formula for this method is:
\[V = 2\pi \int_{a}^{b} \, ext{(radius)} imes ext{(height)} \, dy\]Applied to the problem, we use this method to calculate the volume when the region was revolved around the line \( y = 2 \). By converting expressions into those related to x, it helps set up an integration with respect to x, using the limits from the curves \( y = 0 \) to \( y = 2 \). This technique is effective for regions that would be cumbersome using disk or washer methods.

Integration

Integration is the mathematical process used to find areas or volumes, and it is essential in determining the volume of solids of revolution. By summing up infinitesimally small elements, integration translates the geometric shape into a quantifiable volume.
For volumes of rotations, integration is applied to evaluate the curve's continuous behavior across the defined region. In the disk and washer methods, integration takes the form:
\[\int \\] representing the total area of circular slices. In the shell method, integration tallies up the cylindrical wrap of the region. This exercise involved multiple integrations to solve different parts around axes and lines, converting the region's outlines into understandable integrands that describe their cross-sectional volumes.

Bounded Regions

When dealing with volumes of revolution, the bounded region is the area enclosed by curves that define the shape to be revolved. Understanding the shape and limits of this region is crucial, as they dictate how the solid is formed and the limits of integration.
For the exercise, the bounded region is defined by \( y = \sqrt{x} \), \( y = 2 \), and \( x = 0 \), which intersect at specific points to create a clear outlined area to rotate. It is paramount to visualize this region accurately and to note its boundary limits when setting up integration for finding volumes. In each case, the bounds were identified to guide the integral limits, ensuring the calculated volume corresponds to the genuine shape of the solid created by rotation.