Problem 15 - Calculus: Early Transcendentals

Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. \( y = x^3 \) , \( y = 0 \) , \( x = 1 \) ; about \( x = 2 \)

Step-by-Step Solution

Answer

The volume is \( \frac{10\pi}{3} \).

Step 1: Identify the Region of Interest

The region is bounded by the curve \( y = x^3 \), the line \( y = 0 \), and the vertical line \( x = 1 \). We focus on the interval \( 0 \leq x \leq 1 \) since \( y = x^3 \) only crosses the x-axis at \( x=0 \).

Step 2: Set up the Disk Method

We use the disk method since the solid is obtained by rotating the region around the vertical line \( x = 2 \). For a typical radius, we consider the horizontal distance from any point \( x \) on the curve \( y = x^3 \) to the line \( x = 2 \), which is given by \( R(x) = 2 - x \).

Step 3: Express the Volume Integral

The volume of the solid is calculated using the formula for the volume of a solid of revolution: \[ V = \pi \int_{a}^{b} [R(x)]^2 \, dx \]where \( R(x) = 2 - x \), \( a = 0 \), and \( b = 1 \). Thus, \[ V = \pi \int_{0}^{1} (2-x)^2 \, dx \].

Step 4: Simplify and Integrate the Expression

First, expand the integrand: \((2-x)^2 = 4 - 4x + x^2\). Substitute this into the integral:\( V = \pi \int_{0}^{1} (4 - 4x + x^2) \, dx \).Integrate term by term:\( 4x - 2x^2 + \frac{1}{3}x^3 \bigg|_0^1 \).

Step 5: Evaluate the Definite Integral

Substitute the upper limit \( x = 1 \): \( 4(1) - 2(1)^2 + \frac{1}{3}(1)^3 = 4 - 2 + \frac{1}{3} = \frac{10}{3} \).Substitute the lower limit \( x = 0 \): \( 4(0) - 2(0)^2 + \frac{1}{3}(0)^3 = 0 \).The volume is:\[ V = \pi \left( \frac{10}{3} - 0 \right) = \frac{10\pi}{3} \].

Step 6: Interpret the Result and Sketch

The volume of the solid obtained by rotating the region about \( x = 2 \) is \( \frac{10\pi}{3} \). Sketch the region under \( y = x^3 \) from \( x = 0 \) to \( x = 1 \), the solid as a shape rotated about the line \( x = 2 \), and a typical disk with radius \( 2-x \).

Key Concepts

Disk Method

The Disk Method is a practical technique in calculus used to find the volume of a solid of revolution. Imagine rotating a two-dimensional region around a line to create a three-dimensional shape. To calculate this volume, we consider cross-sections in the form of disks or washers that make up the solid.

In this particular exercise, the region bounded by the curves is rotated around the line \( x = 2 \). Each cross-section perpendicular to the \( x \)-axis resembles a disk. The radius \( R(x) \) of these disks is the horizontal distance from a point on the curve \( y = x^3 \) to the line \( x = 2 \), found by \( R(x) = 2 - x \).

• Disks are circular slices of the solid with varying radius.
• The method implies using a definite integral to sum up these infinitesimally small disks.

Definite Integral

A definite integral is crucial in calculating the total volume when using the Disk Method. It represents the accumulation of infinitely small slices in a bounded interval and gives a precise volume of the solid.

For the exercise given, we express the volume integral as: \[ V = \pi \int_{0}^{1} [R(x)]^2 \, dx \] Here, the integral is taken from \( x = a = 0 \) to \( x = b = 1 \), covering the whole region rotated about \( x = 2 \).

• The integral here figures out the total contribution of all disks between \( a \) and \( b \).
• Each small segment \("dx"\) gives a tiny disk slice whose volume is a part of the sum.

Solid of Revolution

A solid of revolution is a fascinating geometric shape formed when a two-dimensional region is spun around an axis. This process creates a symmetrical three-dimensional object.

In our situation, rotating the region defined by \( y = x^3 \), \( y = 0 \), and \( x = 1 \) about \( x = 2 \) generates such a solid. Visualizing this involves imagining the curve as the edge of the solid which stretches outwards in a circular motion about the chosen axis.

• The horizontal line and curve outline the boundaries of the volume revolved.
• This volume is effectively a set of infinite disks stacked together.

Volume Calculation

Calculating the volume of a solid formed by rotation involves systematic integration. In this exercise, after expanding the integrand of \( (2-x)^2 \), the integral is simplified step-by-step as follows: \[ V = \pi \int_{0}^{1} (4 - 4x + x^2) \, dx \] Each term of the polynomial \((4 - 4x + x^2)\) is integrated separately:

\[ 4x - 2x^2 + \frac{1}{3}x^3 \bigg|_0^1 \]
Evaluating the definite integral from 0 to 1, we substitute and find: \[ V = \pi \left( \frac{10}{3} - 0 \right) = \frac{10\pi}{3} \]

• This tells us the volume of our solid when completely revolved about the axis.
• It's the compact and final step in our solid's volume analysis.