Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

The region bounded by \(y=\sin ^{2}\left(x^{2}\right), y=0\), and \(x=\sqrt{\pi / 2}\) is revolved about the \(y\)-axis. Find the volume of the resulting solid.

Step-by-Step Solution

Answer

The volume is computed using \( V = 2\pi \int_{0}^{1} \sqrt{\arcsin(\sqrt{y})} \, dy \), which typically requires numerical methods to solve.

Step 1: Understand the Problem

We are asked to find the volume of the solid formed by revolving the region bounded by the curve \( y = \sin^2(x^2) \), the line \( y = 0 \), and \( x = \sqrt{\pi/2} \) about the \( y \)-axis. This involves using the method of inverse functions as we revolve it around the \( y \)-axis.

Step 2: Set Up the Inverse Function Form

First, express \( y = \sin^2(x^2) \) in terms of \( x^2 \) by using the inverse trigonometric function. Take the square root of both sides: \( x^2 = \arcsin(\sqrt{y}) \). Since \( x \geq 0 \), \( x = \sqrt{\arcsin(\sqrt{y})} \).

Step 3: Determine Limits for \( y \)

Find the limits for \( y \) by substituting the bounds of \( x \). When \( x = 0 \), \( y = \sin^2(0) = 0 \). When \( x = \sqrt{\pi/2} \), \( y = \sin^2((\sqrt{\pi/2})^2) = \sin^2(\pi/2) = 1 \). Thus, the limits for \( y \) are from 0 to 1.

Step 4: Volume Formula using Shell Method

The volume \( V \) of the solid of revolution using the shell method is given by the integral:\[V = 2\pi \int_{0}^{1} x \cdot h(y)\, dy\]Here, \( h(y) = x = \sqrt{\arcsin(\sqrt{y})} \) represents the lateral surface height at each slice of \( y \).

Step 5: Integrate to Find the Volume

Substitute \( x = \sqrt{\arcsin(\sqrt{y})} \) into the volume integral:\[V = 2\pi \int_{0}^{1} \sqrt{\arcsin(\sqrt{y})} \, dy\]This integral may be complex and require numerical methods or specific software for evaluation. However, in an exam or application context, computational tools or given results for such integrals may be needed.

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that focuses on the concept of integration.
It helps find quantities like areas under curves, totals, and volumes, among other applications. While differential calculus deals with rates of change, integral calculus is concerned with accumulation.

In integral calculus, there are specific techniques to solve different types of integrals. Some of these include:

• Substitution method: used for integrals involving composite functions.
• Integration by parts: helpful for products of functions.
• Partial fraction decomposition: used for rational functions.

The integral calculus plays an essential role in finding volumes of solids, especially when dealing with areas bounded by curves.
This involves setting up integral expressions that account for the rotating or stacking of the area to determine the volume.

Volume of Solids of Revolution

The volume of solids of revolution involves rotating a region or area about a specific axis to create a 3-dimensional solid.
In the given exercise, the region rotated about the y-axis creates a shape known as a solid of revolution. Several methods exist to calculate the volumes of such solids, but the shell method was used here.

The shell method involves slicing the solid into cylindrical shells. To compute the volume, you:

• Consider a small strip of the region and rotate it around the axis.
• Calculate the lateral surface area of the shell formed by the strip.
• Integrate this surface area over the interval to find the total volume.

The formula for the volume using the shell method is:\[V = 2\pi \, \int_{a}^{b} \text{radius} \, \times \, \text{height} \, dx\]In the exercise:- **Radius:** Represents the horizontal distance from the strip to the axis of rotation.- **Height:** Given by the inverse function in terms of the rotated variable.This technique allows us to find volumes of complex shapes by summing up these shells.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus whenever we want to solve problems involving angles and trigonometric expressions.
These functions reverse the actions of the standard trigonometric functions.

For example, the inverse sine function, written as \( \arcsin(y) \), is used when we know the sine value and want to determine the corresponding angle.

These functions have specific ranges over which they are defined. For instance:

• \( \arcsin(y) \) returns values between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
• \( \arccos(y) \) gives angles from 0 to \(\pi\).
• \( \arctan(y) \) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

In the exercise, the inverse function \( \arcsin(\sqrt{y}) \) helps express \(x\) in terms of \(y\) when rotated around the y-axis.
Understanding how these functions work allows us to manipulate expressions in integrals and solve for variables needed to determine the overall outcome in calculus problems.