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Chapter 8: Problem 56

Finding volume Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve \(y=e^{-x},\) and the line \(x=1\) a. about the \(y\) -axis. b. about the line \(x=1\)

Short Answer

Expert verified
a. Volume around y-axis: \(2\pi(1-e^{-1})\). b. Volume around x=1: \(\pi(\frac{3}{2}-\frac{1}{2}e^{-2})\).

Step by step solution

01

Understand the Problem

We have a region formed by the curve \(y = e^{-x}\), the vertical line \(x = 1\), and the coordinate axes in the first quadrant. Our task is to find the volume of the solid created by revolving this region around two different axes.
02

Set up the integral for revolving around the y-axis

When revolving around the \(y\)-axis, we can use the method of cylindrical shells. The height of the shell is \(y = e^{-x}\). The radius of the shell is \(x\). We integrate from \(x=0\) to \(x=1\). The volume is given by \( V = \int_0^1 2\pi x e^{-x} \, dx \).
03

Evaluate the integral for the y-axis

Compute the integral \( V = 2\pi \int_0^1 x e^{-x} \, dx \). Using integration by parts, let \(u = x\) and \(dv = e^{-x} \, dx\). Then, \(du = dx\) and \(v = -e^{-x}\). The integral becomes \(-2\pi \left[x e^{-x} \bigg|_0^1 - \int_0^1 -e^{-x} \, dx\right]\). Simplify and evaluate to find \( V = 2\pi (e^{-1} - e^{-1} + 1) = 2\pi (1 - e^{-1})\).
04

Set up the integral for revolving around x=1

Revolving around \(x=1\) uses the method of washers or discs. The outer radius is \(1\), and the inner radius is \(1-x\). The area of a washer at a point \(y\) is \(\pi (1)^2 - \pi (e^{-x})^2\). The volume is \( V = \pi \int_0^1 (1 - (e^{-x})^2) \, dx\).
05

Evaluate the integral for x=1

Compute the integral \( V = \pi \int_0^1 (1 - e^{-2x}) \, dx \). Break it into two integrals: \(\pi [\int_0^1 1 \, dx - \int_0^1 e^{-2x} \, dx]\). The solution becomes \(\pi [1 - (-\frac{1}{2}e^{-2x} \bigg|_0^1)] = \pi [1 - (-\frac{1}{2}(e^{-2} - 1))]\), which simplifies to \(\pi (1 + \frac{1}{2} - \frac{1}{2}e^{-2})\).

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Key Concepts

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

Cylindrical Shells
Cylindrical shells are a powerful method used in finding the volume of a solid of revolution, especially when the region is revolved around the y-axis.

This method involves imagining the solid as consisting of many thin, hollow tubes or shells.
Each shell can be visualized as having:
  • A height equivalent to the function value, here given by \(y = e^{-x}\).
  • A radius equal to the horizontal distance from the y-axis, which is simply \(x\).
By integrating these individual shell volumes along the x-axis, from \(x = 0\) to \(x = 1\), we approximate the entire volume of the solid.
The formula for the volume using cylindrical shells is given by the integral \( V = \int_0^1 2\pi x e^{-x} \, dx \), where \(2\pi x\) is the circumference of the shell and \(e^{-x}\) is its height.
Integration by Parts
Integration by parts is a crucial technique used to solve integrals that cannot be easily computed.
It is particularly handy for products of functions, such as in the integral \( \int x e^{-x} \, dx \).

The method is based on the product rule for differentiation, and the formula used is:
\[ \int u \, dv = uv - \int v \, du \]
This technique helps transform a complex integral into simpler parts that are easier to manage.
  • Choose \(u = x\), which makes \(du = dx\).
  • Let \(dv = e^{-x} \, dx\), allowing us to find \(v = -e^{-x}\).
By plugging these into the integration by parts formula, we can compute the required integral to find the volume when using cylindrical shells around the y-axis.
Washers or Discs
When revolving a region around a line other than the y-axis, such as \(x=1\), the washers or discs method becomes highly useful.
This approach divides the solid into flat disc-like shapes, resembling washers, each with:
  • An outer radius corresponding to the maximum distance from the axis of revolution.
  • An inner radius derived from the curve limiting the solid, here \(y = e^{-x}\).
For this problem, the outer radius is \(1\), and the inner radius is \(1-x\).
The volume of each washer or disc is the difference between the area of the larger and smaller circles.
The integral for the volume is then \( V = \pi \int_0^1 (1 - (e^{-x})^2) \, dx \), capturing the region's solid volume.
First Quadrant
The first quadrant in the Cartesian coordinate system is the region where both x and y are positive.

This is important for understanding the boundaries and limits of integration because our region of interest must lie entirely within this section.
  • The first quadrant is defined by the x-axis (where \(y = 0\)) and the y-axis (where \(x = 0\)).
  • Our region includes the curve \(y = e^{-x}\) and is bounded by \(x = 1\), ensuring it remains in the first quadrant.
Thus, the limits of integration, \(x = 0\) to \(x = 1\), and the function values \(y = e^{-x}\), all operate within the positive x and y values of the first quadrant, guiding how we set up our volumes for revolution calculations.

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