Problem 14 - Thomas Calculus

Question

Use the shell method to find the volumes of the solids generated by re- volving the regions bounded by the curves and lines about the \(y\)-axis. Let \(g(x)=\left\\{\begin{array}{ll}{(\tan x)^{2} / x,} & {0< x \leq \pi / 4} \\\ {0,} & {x=0}\end{array}\right.\) a. Show that \(x g(x)=(\tan x)^{2}, 0 \leq x \leq \pi / 4\) b. Find the volume of the solid generated by revolving the shaded region about the \(y\)-axis.

Step-by-Step Solution

Answer

The volume of the solid is \(2\pi - \frac{\pi^2}{2}\).

Step 1: Understand the Function

The given function is a piecewise function defined as \[g(x) = \begin{cases} (\tan x)^2 / x, & 0 < x \leq \pi/4 \ 0, & x = 0 \end{cases}\]We will work within the interval \(0 \leq x \leq \pi/4\). For \(x > 0\), clearly, \[x \cdot g(x) = x \cdot \left(\frac{(\tan x)^2}{x}\right) = (\tan x)^2\].

Step 2: Setting up the Shell Method

The shell method formula for finding the volume of a solid of revolution about the \(y\)-axis is:\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]Given that \(x \cdot g(x) = (\tan x)^2\), substitute into the shell method formula:\[ V = 2\pi \int_{0}^{\pi/4} x \cdot (\tan x)^2 \, dx = 2\pi \int_{0}^{\pi/4} (\tan x)^2 \, dx \]

Step 3: Using a Trigonometric Identity

Substitute the identity \((\tan x)^2 = (\sec x)^2 - 1\) into the integral:\[ V = 2\pi \int_{0}^{\pi/4} ((\sec x)^2 - 1) \, dx \]Now, separate the integral:\[ V = 2\pi \left( \int_{0}^{\pi/4} (\sec x)^2 \, dx - \int_{0}^{\pi/4} 1 \, dx \right) \]

Step 4: Evaluating the Integrals

Evaluate each integral separately:- For \(\int (\sec x)^2 \, dx\), the antiderivative is \(\tan x\). Thus, \[\int_{0}^{\pi/4} (\sec x)^2 \, dx = \tan x \Bigg|_{0}^{\pi/4} = \tan(\pi/4) - \tan(0) = 1 - 0 = 1 \]- For \(\int 1 \, dx\), the antiderivative is \(x\). Thus, \[\int_{0}^{\pi/4} 1 \, dx = x \Bigg|_{0}^{\pi/4} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

Step 5: Solve for the Volume

Substitute the evaluated integrals back into the expression:\[ V = 2\pi \left( 1 - \frac{\pi}{4} \right) \]

Step 6: Simplify the Expression

Calculate the simplified volume:\[ V = 2\pi \left( 1 - \frac{\pi}{4} \right) = 2\pi - \frac{\pi^2}{2} \]The final volume of the solid is \(2\pi - \frac{\pi^2}{2}\).

Key Concepts

Shell Method

The Shell Method is a powerful technique used in calculus to find the volume of solids of revolution. This method is particularly useful when the solid is generated by revolving a region around either the x-axis or y-axis. The basic principle relies on adding up the volumes of numerous cylindrical shells, formed in rotation around an axis.

When using the shell method to find the volume around the y-axis, the formula is:

• \( V = 2\pi \, \int_{a}^{b} x \cdot f(x) \, dx \)

Here, \( x \) represents the radius of each cylindrical shell, and \( f(x) \) is the height of the shell.

The integration process involves; solving for individual volumes, from the beginning to the end of the specified interval (from \( a \) to \( b \)). This method is especially helpful when vertical slices of the region are easier to integrate, such as when revolving curves around non-horizontal axes.

Piecewise Functions

Piecewise functions are functions categorized into multiple "pieces", each defined over a particular segment of the domain.

They are significant when a single expression cannot adequately describe the function across its entire domain.

In this case, piecewise functions are written using:

• A set of conditions or intervals.
• Corresponding expressions for each condition.

The main aim is to provide accurate representation over different intervals. For example, consider the piecewise function given by:
\[ g(x) = \begin{cases} (\tan x)^2 / x, & 0 < x \leq \pi/4 \ 0, & x = 0 \end{cases} \]
This function highlights two intervals; the function behaves one way for \( 0 < x \leq \pi/4 \), and is zero at \( x = 0 \).

Understanding how piecewise functions behave differently across intervals is crucial, especially in problems involving abrupt changes in conditions, like those often found at boundaries.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable for which both sides of the equation are defined.

These identities are helpful in simplifying and evaluating integrals involving trigonometric functions. One commonly used identity is:

• \((\tan x)^2 = (\sec x)^2 - 1\)

This identity was essential in the step-by-step solution example. It helped simplify the integral \( \int (\tan x)^2 \, dx \) into a more manageable form, making it easier to solve. Breaking down complicated expressions into simpler terms speeds up the problem-solving process and aids in finding solutions more efficiently.

Solid understanding of trigonometric identities allows students to manipulate and transform equations, making them more accessible.

Integral Calculus

Integral calculus is a primary branch of calculus focusing on finding the total accumulation of quantities and the areas under and between curves.

The main operations in integral calculus include finding antiderivatives (indefinite integrals) and evaluating definite integrals which are related to area and accumulation:

• The indefinite integral represents a family of functions and includes a constant of integration \( C \).
• The definite integral, evaluated over an interval \( [a, b] \), represents the exact area under the curve between these points.

To solve problems found within an interval, each integral is evaluated separately, often involving techniques like substitution and algebraic transformation.

In practical applications, like the problem of volume calculation by revolving around an axis, setting up proper integrals is critical for successful problem-solving. By applying integral calculus, one can calculate not just positions or rates of change, but also tangible quantities, like the volume of revolved solids.