Problem 35 - 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 \(y=\sqrt{x}\) and \(y=x^{2} / 8\) about $$ \text { a. the }x \text { -axis } \quad \text { b. the y-axis} $$

Step-by-Step Solution

Answer

a: \(\frac{128\pi}{5}\), b: \(\frac{48\pi}{5}\)

Step 1: Set up the problem

We need to find the volumes of two solids of revolution generated by revolving the region bounded by the curves \(y = \sqrt{x}\) and \(y = \frac{x^2}{8}\). In part (a), the region revolves around the \(x\)-axis, while in part (b), it revolves around the \(y\)-axis.

Step 2: Find the intersection points

First, we find the intersection points by setting \(\sqrt{x} = \frac{x^2}{8}\). Solving \(x = \frac{x^4}{64}\), simplifying gives \(x^4 - 64x = 0\). Factoring, \(x(x^3 - 64) = 0\) gives solutions \(x = 0\) and \(x = 4\). These are the bounds for integration.

Step 3: Set up the integral for part (a)

For revolving around the \(x\)-axis, we use the washer method. The volume \(V\) is given by: \[ V = \pi \int_{0}^{4} \left(\left(\sqrt{x}\right)^2 - \left(\frac{x^2}{8}\right)^2\right)\, dx. \] Simplifying, \(V = \pi \int_{0}^{4} (x - \frac{x^4}{64}) \, dx\).

Step 4: Evaluate the integral for part (a)

We compute the integral: \(\pi \int_{0}^{4} (x - \frac{x^4}{64}) \, dx = \pi \left[\frac{x^2}{2} - \frac{x^5}{320}\right]_{0}^{4}\). Calculating the definite integral, \(\pi \left[\frac{16}{2} - \frac{1024}{320}\right] = \pi \cdot \frac{128}{5}\), so the volume is \(\frac{128\pi}{5}\).

Step 5: Set up the integral for part (b)

For revolving around the \(y\)-axis, change variables. Solve \(y = \sqrt{x}\) giving \(x = y^2\), and \(y = \frac{x^2}{8}\) giving \(x = 2\sqrt{2y}\). Using cylindrical shells, the volume is: \[ V = 2\pi \int_{0}^{2} (y^2 - 2\sqrt{2y})y \, dy. \] This simplifies to \(2\pi \int_{0}^{2} (y^3 - 2y^{3/2}) \, dy\).

Step 6: Evaluate the integral for part (b)

Compute \(2\pi \int_{0}^{2} (y^3 - 2y^{3/2}) \, dy = 2\pi \left[\frac{y^4}{4} - \frac{4y^{5/2}}{5}\right]_{0}^{2}\). Calculating gives \(2\pi \left[4 - \frac{32}{5}\right] = \frac{48\pi}{5}\). Thus, the volume for part (b) is \(\frac{48\pi}{5}\).

Key Concepts

Washer Method

When we revolve a region around an axis, and the region is bounded by two curves, the washer method is often a good choice. This method calculates the volume by subtracting the volume of a solid cylinder from another larger one, which creates a hollow, washer-like structure in between.
The formula for volume using the washer method is:

• The outer radius, which is the distance from the axis of revolution to the outer curve, and
• The inner radius, measured from the axis to the inner curve.

For volume calculation, use:\[ V = \pi \int_0^a \left([\text{outer radius}]^2 - [\text{inner radius}]^2\right) \, dx\]For example, revolving curves such as \(y=\sqrt{x}\) and \(y=\frac{x^2}{8}\) around the x-axis, you'll set the outer and inner radii based on distance from axis to curves, providing you the bounds and information for setting up the integral.

Cylindrical Shells

The cylindrical shells method alternates between the approaches of parallel and perpendicular integration to find the volume of a solid of revolution. It's particularly useful when revolving around an axis that is not parallel to the curves being integrated.
This method is well-suited for cases where washers or disks would be complex due to simple algebraic manipulations. Use the formula:

• The height of the shell is the difference between the functions along the same axis.
• The radius is the distance from the chosen axis to the average point of the shell.

The formula for this method is:\[ V = 2 \pi \int_a^b [\text{radius}][\text{height}] \, dy\]In our example, for the problem revolving around the y-axis, cylindrical shells are efficient. Revolve functions like \(y=\sqrt{x}\) converting them appropriately to help set up the integrals.

Volume Integration

Volume integration is the process of calculating the volume of a 3-dimensional solid using calculus. It involves integrating cross-sectional areas over a range.
To perform volume integration accurately:

• Choose the right method, either washers, disks, or cylindrical shells, depending on problem requirements.
• Identify bounds of integration using clear intersections and then set up the integral accordingly.

Volume integration is crucial for accurately determining the solid's entire volume formed by rotating any given region around an axis. In practice, integrate the area functions derived from the given equation over the appropriate range.

Intersection Points

Before setting up your integrals for volume calculation, find intersection points of the functions. Intersection points indicate where the functions cross each other and determine the limits of integration.
To find these points:

• Set the functions equal, for example, \(\sqrt{x} = \frac{x^2}{8}\).
• Solve the resulting equation for x or y, to see where the curves intersect.

In our example, these intersection points occur at \(x = 0\) and \(x = 4\), which provide crucial information about the solid's bounds and thus guide integral setup.

Definite Integral

A definite integral represents the area under a path or region on a graph between two points, commonly used in finding areas and volumes.
In volumetric problems:

• Definite integrals provide the exact volume between intersections.
• Perform integration over these set bounds to find solid areas or differences between curves.

When calculated, the result is the solid's volume in numerical form. For instance, the definite integral of \(\pi \int_0^4 (x - \frac{x^4}{64}) \, dx\) yielded the volume \(\frac{128\pi}{5}\). Such calculations are standardized steps in solid volume determination using calculus.