Problem 130 - Calculus Volume 1

Question

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and y = 0 around the x-axis. $$y=\sqrt{1-x^{2}}, x=0, \text { and } x=1$$

Step-by-Step Solution

Answer

\( \frac{2\pi}{3} \)

Step 1: Understand the Problem

The region is bounded by the curve \( y = \sqrt{1-x^2} \), the x-axis \( (y=0) \), and the vertical lines \( x = 0 \) and \( x = 1 \). We need to find the volume by rotating this region around the x-axis.

Step 2: Set Up the Volume Formula with Shell Method

The formula for the shell method is given by \( V = 2\pi \int_{a}^{b} x \, h(x) \, dx \), where \( h(x) \) is the height of the shell. For rotation around the x-axis, \( h(x) = \sqrt{1-x^2} \).

Step 3: Determine the Limits of Integration

The limits of integration \( a \) and \( b \) are determined by the region boundaries along the x-axis: \( x = 0 \) and \( x = 1 \). Therefore, we have \( a = 0 \) and \( b = 1 \).

Step 4: Integrate to Find the Volume

Substitute the function and limits of integration into the shell formula: \[ V = 2\pi \int_{0}^{1} x \sqrt{1-x^2} \, dx \] Use the substitution method with \( x = \sin(\theta) \), so \( dx = \cos(\theta) \, d\theta \). The integral becomes: \[ V = 2\pi \int_{0}^{\pi/2} \sin(\theta) \cdot \cos^2(\theta) \, d\theta \] Simplify: \[ V = 2\pi \int_{0}^{\pi/2} \sin(\theta) (1 - \sin^2(\theta)) \, d\theta \] Divide into two integrals: \[ V = 2\pi \left( \int_{0}^{\pi/2} \sin(\theta) \, d\theta - \int_{0}^{\pi/2} \sin^3(\theta) \, d\theta \right) \] Solve these integrals to find the volume.

Step 5: Solve Integrals Separately

Calculate each part of the final expression:1. \( \int_{0}^{\pi/2} \sin(\theta) \, d\theta = 1 \)2. \( \int_{0}^{\pi/2} \sin^3(\theta) \, d\theta = \frac{2}{3} \)Substitute back into the expression: \[ V = 2\pi \left( 1 - \frac{2}{3} \right) = 2\pi \cdot \frac{1}{3} = \frac{2\pi}{3} \]

Step 6: Conclude the Solution

The volume of the solid generated by rotating the given region around the x-axis is \( \frac{2\pi}{3} \).

Key Concepts

Volume of Revolution

When we talk about finding the volume of a revolution, we are referring to the volume of a 3D object that is created when a 2D region is rotated around an axis. In this particular problem, we're revolving the region between the curve \( y = \sqrt{1-x^2} \), the x-axis, and the lines \( x = 0 \) and \( x = 1 \) around the x-axis to form a solid. Imagine this as the surface of a half-circle being spun around a rod to generate a symmetrical shape. To calculate this volume, we can use methods like the washer or disk method, but here, we're using the shell method. This method is especially useful when the region is easy to describe in terms of height as a function of one variable. Understanding this concept is essential when handling two-dimensional areas extending into three dimensions through rotation.

Integral Calculus

Integral calculus is a fundamental tool for calculating areas and volumes. The basis of integral calculus is the integral, which accumulates quantities over a particular interval on a curve or a surface. When finding volumes or areas, figuring out the integral of a function helps determine the size of the region.In this exercise, we use the integral \( \int_{a}^{b} x \sqrt{1-x^2} \, dx \) from the shell method to accumulate the infinite sum of cylindrical shells' volumes around the x-axis. The result from this integration provides the solid's volume. Thus, integral calculus serves not only in evaluating the size of geometrical shapes but also in understanding the properties and behaviors of functions involved in generating these shapes. Learning to properly set up and evaluate these integrals is a powerful skill that extends beyond simple volumes to areas such as physics, engineering, and beyond.

Substitution Method

The substitution method can simplify complex integrals, often making them easier to solve. It works by changing variables to transform the integrand into a simpler form. In the given problem, using substitution helps us solve the integral \( \int_{0}^{1} x \sqrt{1-x^2} \, dx \).Here, we substitute \( x \) with \( \sin(\theta) \), which transforms the integral into a trigonometric form: \( \int_{0}^{\pi/2} \sin(\theta) \cdot \cos^2(\theta) \, d\theta \). This transformation is beneficial because integrating trigonometric expressions typically simplifies the problem. Understanding when and how to apply values in a substitution, especially with trigonometric simplifications, is a valuable part of problem-solving in calculus. It allows tackling more complex integrals with ease, cultivating a more profound computational intuition.