Problem 49 - 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 a. the \(x\)-axis. \(\quad\) b. the \(y\)-axis. c. the line \(y=2 . \quad\) d. the line \(x=4\)

Step-by-Step Solution

Answer

a) \(8\pi\), b) \(6\pi\), c) \(\frac{8}{3}\pi\), d) \(\frac{512}{15}\pi\)

Step 1: Sketch the Region

First, we need to understand the region enclosed by the curve and lines. The curve given is \(y = \sqrt{x}\), which is a parabola opening to the right. We also have the horizontal line \(y = 2\) and the vertical line \(x = 0\). The region of interest is to the right of \(x = 0\), below \(y = 2\), and above the curve \(y = \sqrt{x}\).

Step 2: Volume Around the x-axis (Part a)

To find the volume when the region is revolved around the \(x\)-axis, we use the disk method. The radius of a typical disk is \(y\), and its thickness is \(dx\). The limits of integration are from \(x = 0\) to the point where \(\sqrt{x} = 2\), which is \(x = 4\). Thus, the volume \(V\) is: \[ V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx = \pi \int_{0}^{4} x \, dx \] \[ V = \pi \left[ \frac{x^2}{2} \right]_{0}^{4} = \pi \times \frac{16}{2} = 8\pi \]

Step 3: Volume Around the y-axis (Part b)

For revolution around the \(y\)-axis, we use the shell method. The height of a shell is \(2 - \sqrt{x}\) (from \(y=2\) to \(y=\sqrt{x}\)) and its radius is \(x\). The limits are from \(y = 0\) to \(y = 2\). We first express \(x\) in terms of \(y\), yielding \(x = y^2\). The volume \(V\) is: \[ V = 2\pi \int_{0}^{2} y (2-y^2) \, dy \] \[ V = 2\pi \left[ 2y^2/2 - y^4/4 \right]_{0}^{2} \] \[ V = 2\pi \left[ 4 - 4/4 \right] = 2\pi \times 3 = 6\pi \]

Step 4: Volume Around the line y=2 (Part c)

For revolution about the line \(y = 2\), we use the washer method. The outer radius is \(2 - 0 = 2\) and the inner radius is \(2 - \sqrt{x}\). Using the limits from \(x = 0\) to \(x = 4\), the volume \(V\) is: \[ V = \pi \int_{0}^{4} \left( (2)^2 - (2-\sqrt{x})^2 \right) \, dx \] Solving yields: \[ V = \pi \int_{0}^{4} (4 - (4 - 4\sqrt{x} + x)) \, dx = \pi \int_{0}^{4} (4\sqrt{x} - x) \, dx \] \[ V = \pi \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{2} \right]_{0}^{4} = \frac{32}{3}\pi - 8\pi = \frac{8}{3}\pi \]

Step 5: Volume Around the line x=4 (Part d)

We use the shell method for revolution around \(x = 4\). The radius of a shell is \(4 - x\) and the height is \(2 - \sqrt{x}\). The limits are from \(x = 0\) to \(x = 4\): \[ V = 2\pi \int_{0}^{4} (4-x)(2-\sqrt{x}) \, dx \] After expanding and solving, we find: \[ V = 2\pi \left( 8x - 4x^{3/2} - 2x^2/2 + x^{5/2}/5 \right)_{0}^{4} \] \[ = 2\pi \left( 32 - 64/3 + 16 - 32/5 \right) \] \[ = 2\pi \left( \frac{240 - 64 \times 5/3 - 32/5}{15} \right) = \frac{512}{15}\pi \]

Key Concepts

Disk Method

The disk method is a technique used to determine the volume of a solid of revolution when an area is revolved around an axis. Imagine stacking an infinite number of tiny disks or circles, each having a small thickness. The volume is then the sum of the volumes of these small disks.

This method applies well when revolving a region around the x-axis or y-axis, where the solid forms a series of disks perpendicular to the axis of rotation. To find the volume, you need:

• The radius of the disk, which is the distance from the axis to the outer edge of the region.
• The thickness of the disk, which is a small change along the axis, noted by the differential element, such as 'dx' or 'dy'.

For example, revolving around the x-axis implies a disk's radius equals the function value at a certain x, i.e., the height of the curve from the axis. If revolving around the y-axis, consider the radius determined by the x-values.

Shell Method

The shell method focuses on visualizing the solid as a series of cylindrical shells rather than disks. You will use the shell method when a region is revolved around a vertical line like the y-axis or a horizontal line, not through the region's immediate boundary.

The shell's radius is the distance from the shell to the axis of rotation, and its height is the function value or the distance between two functions producing the bounded region. To apply this method:

• Identify the shell radius and height.
• The thickness of the shell is the small differential, such as 'dx' or 'dy'.

Generally, for a region around the y-axis, the shell radius is x, and the shell height depends on the functional relationship along y. This technique becomes useful for rotations involving horizontal bounds or when revolving around lines parallel to the x-axis or y-axis.

Washer Method

The washer method extends the idea of the disk method by accounting for a hollow center when revolving an area with an inner boundary. Imagine each part of the solid as a washer, essentially a disk with a hole in the middle.

This method is essential when the solid involves revolutions around axes not touching the area or when the region has a clear inner boundary. The washing machine involves:

• An outer radius, determined by the outermost boundary of the region from the axis.
• An inner radius, which is the distance between the axis and the inner boundary of the region.

The final volume is determined by subtracting the inner cylinder's volume from the outer cylinder's volume across the interval or region of integration. This gives a hollowed-out effect akin to many washers stacked together. This method is particularly helpful when dealing with more complex shapes.

Integration

Integration represents the mathematical tool used to sum up infinitely small quantities, providing a vast variety of applications beyond calculating simple areas under curves. In the context of volume, integration helps add up all the tiny volumes of disks, shells, or washers formed by revolution.

For volume calculations:

• Integration aggregates volumes over the designated interval.
• The integrand function models the geometric shape's cross-sections.

By integrating, we mathematically formalize the accumulation of small parts. When dealing with volume, consider definite integrals with respect to the function's variable and set appropriate limits to cover the complete solid region being studied. Integration allows precise calculation from these methods, materializing the theoretical volume into a finite number.

Definite Integrals

Definite integrals are a specific type of integral used to calculate the exact accumulation (such as area or volume) over an interval. When applying definite integrals in volume calculations, they provide the total accumulation across a set of bounded values.

In the context of solids of revolution, definite integrals are pivotal because:

• They assure absolute values are computed, covering changes in shape.
• Each integral has clear bounds, dictated by the limits of the region for easy assessment.

With definite integrals, you compute the total volume or area from set bounds without approximation guesswork. The use of definite integrals implies that each section within the interval contributes accurately to the final calculated measure. Definite integrals thus represent certainty in accumulation, vital for determining the revolved solid's complete volume on a fixed interval using any of the described methods.