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Chapter 5: Problem 15

Sketch the region \(R\) bounded by \(y=1 / x^{3}, x=1, x=3\), and \(y=0 .\) Set up (but do not evaluate) integrals for each of the following. (a) Area of \(R\) (b) Volume of the solid obtained when \(R\) is revolved about the \(y\) -axis (c) Volume of the solid obtained when \(R\) is revolved about \(y=-1\) (d) Volume of the solid obtained when \(R\) is revolved about \(x=4\)

Short Answer

Expert verified
Set up necessary integrals, but do not evaluate them.

Step by step solution

01

Plot the functions and understand the boundaries

Sketch the curve of the function given: \( y = \frac{1}{x^3} \), which is a hyperbola. Identify the vertical boundaries \( x = 1, x = 3 \), and the horizontal boundary \( y = 0 \), which is the x-axis. The region \( R \) is now clearly visible as it is bounded by these lines and the curve.
02

Setting up the integral for the area of region R

To find the area, set up the definite integral across the x-bounds where the curve lies. The integral for the area is: \[ A = \int_{1}^{3} \frac{1}{x^3} \, dx. \] This represents the accumulation of infinitesimally thin vertical slices from \( x=1 \) to \( x=3 \).
03

Volume of solid revolved about the y-axis

Using the shell method due to the axis of rotation being vertical, the volume \( V \) can be set up as follows: \[ V = 2\pi \int_{1}^{3} x \frac{1}{x^3} \, dx = 2\pi \int_{1}^{3} \frac{1}{x^2} x \, dx. \] Here, \( x \times \text{{height}} \times \text{{thickness}} \) is used to account for cylindrical shells with height \( \frac{1}{x^3} \).
04

Volume of solid revolved about y = -1

Shift the function down by 1 unit to set up the washer method integral around the line \( y = -1 \). The new height from \( y = -1 \) to \( y = \frac{1}{x^3} \) is \( \frac{1}{x^3} + 1 \). Set the volume integral: \[ V = \pi \int_{1}^{3} ((\frac{1}{x^3} + 1)^2 - 1^2) \, dx. \] Use outer radius minus inner radius formulation for washers.
05

Volume of solid revolved about x = 4

Since the revolving axis \( x=4 \) is vertical, use the shell method. The radius from the shell will be \( 4-x \) and the height will remain \( \frac{1}{x^3} \). Set up the integral: \[ V = 2\pi \int_{1}^{3} (4-x) \frac{1}{x^3} \, dx. \] Integrate over the thickness \( dx \) from \( x=1 \) to \( x=3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area Calculation
When dealing with calculus problems, area calculation is a fundamental skill. In this exercise, we calculated the area of region \( R \) which is bound by the function \( y = \frac{1}{x^3} \) and the lines \( x = 1 \), \( x = 3 \), and \( y = 0 \). To set up the definite integral for area:
  • First, identify the region enclosed by these curves and lines, which lies between \( x = 1 \) and \( x = 3 \).
  • The curve is above the \( x \)-axis over this region.
  • The integral \( A = \int_{1}^{3} \frac{1}{x^3} \, dx \) adds up the infinitesimally small vertical slices from \( x = 1 \) to \( x = 3 \).
This method essentially sums up the areas of infinitely thin rectangles under the curve \( y = \frac{1}{x^3} \). This approach provides the total area of \( R \) between the specified x-values.
Volume of Revolution
Volume of revolution refers to the volume of a 3-dimensional shape formed by rotating a 2-dimensional region around a specified axis. Each way you revolve the region, you form a different solid:
  • Revolving \( R \) around the \( y \)-axis (vertical axis) involves creating concentric cylindrical shells.
  • Rotation around \( y = -1 \) means the axis is parallel to the \( x \)-axis, creating a series of washers stacked along the rotation path.
  • Revolving about \( x = 4 \) similarly involves cylindrical shells, but with a different radius.
Different methods like the shell method or washer method help calculate these volumes accurately, depending on the axis of rotation. Each setup follows the same basic idea: integrate across the region to sum the volume of the infinitesimal parts.
Definite Integrals
Definite integrals play a crucial role in calculating areas and volumetric properties of regions. They calculate the total accumulation of quantities, like area under a curve or the volume of revolution:
  • For areas, they sum the vertical slices or washers/rings formed between two x-bounds or y-bounds.
  • For volumes, they account for 3D shapes formed by the revolution of 2D areas across a given axis.
The setup of these integrals needs careful consideration of bounds and functions. For instance, the definite integral \( \int_{1}^{3} \frac{1}{x^2} x \, dx \) when using the shell method measures the entire volume by adding the effects of infinitely small cylindrical shells.
Shell Method
The shell method is a technique used to find the volume of a solid of revolution when the axis of rotation is vertical. This method is particularly useful when rotating a region around a y-axis or any line parallel to it. Here's how it works:
  • You imagine the region divided into thin cylindrical shells. Each shell has a small thickness \( dx \).
  • The radius of each shell is the distance from the axis of rotation to the slice. For instance, when revolving around \( x = 4 \), the radius is \( 4-x \).
  • The height of each shell is given by the function value, here \( \frac{1}{x^3} \).
  • The formula becomes \( V = 2\pi \int_{1}^{3} \text{radius} \times \text{height} \, dx \).
The shell method integrates across the interval [1,3] to account for the full volume created by all shells, leading to the complete solid's volume by considering each shell individually and summing across.

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Most popular questions from this chapter

Find the volume of the solid generated by revolving about the \(y\) -axis the region bounded by the line \(y=4 x\) and the parabola \(y=4 x^{2}\).

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