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Chapter 7: Problem 39

Finding the Volume of a Solid In Exercises\(37 - 40 ,\) find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(x\) -axis. Verify your results using the integration capabilities of a graphing utility. $$y = e ^ { x - 1 } , \quad y = 0 , \quad x = 1 , \quad x = 2$$

Short Answer

Expert verified
The volume of the solid generated is \(0.5\pi(e^2 - 1)\).

Step by step solution

01

Set up the integral for the volume

By using the disk method for finding the volume of a solid of revolution, set up the integral according to the formula \(V = \pi \int_{a}^{b}[f(x)]^2 dx\). Here, \(f(x) = e ^ { x - 1 }\), a is 1, and b is 2. So the integral becomes \(V = \pi \int_{1}^{2}[e ^ { x - 1 }]^2 dx\).
02

Simplify the integrand

The square of \(e ^ { x - 1 }\) is \(e ^ { 2(x - 1) }\), so the integral becomes \(V = \pi \int_{1}^{2} e ^ { 2(x - 1)} dx\).
03

Make a substitution

Let \(u = 2(x - 1)\). The derivative of \(u\) is \(du = 2dx\) and therefore \(dx = du / 2\). In the new limits, when \(x = 1\), \(u = 0\), and when \(x = 2\), \(u = 2\). So the integral, now in terms of \(u\), becomes \(V = \pi \int_{0}^{2} e ^ { u } (du / 2)\).
04

Evaluate the integral

The antiderivative of \(e^u\) is \(e^u\). Perform the integration: \(V = \pi [0.5 * e^u ]_{0}^{2}\). This evaluates to \(\pi * 0.5 * (e^2 - e^0) = 0.5\pi(e^2 - 1)\).
05

Simplify the final answer

The final answer for the volume simplifies to \(V = 0.5\pi(e^2 - 1)\)

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Key Concepts

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

Disk Method
The disk method is a powerful tool to calculate the volume of a solid of revolution. Imagine revolving a region around an axis, creating a 3D shape. The disk method works by slicing this solid into thin circular disks.
Each disk's volume is then calculated, and the volumes are summed to get the total volume. Mathematically, this is expressed using an integral, specifically:

\[ V = \pi \int_{a}^{b} [f(x)]^2 dx \]
Here, \( V \) is the volume, \( [f(x)]^2 \) is the radius of a disk squared, and \( dx \) represents the thickness of each disk.
  • Identify the function \( f(x) \) that describes the boundary of your region.
  • Set the limits of integration, \( a \) and \( b \), which relate to the bounds of the region you are revolving.
  • Compute the integral to find the total volume.
The result is a formula that gives you the entire volume of the solid. It's important to ensure the function being squared corresponds to the correct distance from the axis of rotation.
Integral Calculus
Integral calculus is a branch of mathematics focused on the concept of accumulation and areas under curves. When it comes to finding volumes of solids of revolution, integration is the method used to sum an infinite number of infinitesimal quantities, like those little disks from the disk method.

To solve integrals like \( \int_{1}^{2} e^{2(x-1)} \, dx \), you might use substitution for simplification:
  • Set \( u = 2(x - 1) \).
  • Find the derivative \( du = 2 \, dx \), hence \( dx = du / 2 \).
  • Change the limits accordingly, adjusting them to \( u \) which may simplify further integration.
In this exercise, substitution helps transform the original variable into one that is easier to integrate. After integration, evaluate within the limits to find the exact volume.
Graphing Utility
A graphing utility can be an invaluable tool in calculus for visualizing functions, verifying solutions, and analyzing complex integrals. Modern calculators or software allow you to quickly compute integrals numerically, which is particularly useful in checking manually derived solutions.

Here's how a graphing utility can help with volume of revolution problems:
  • Input the function and set your integral limits.
  • Use the built-in functions to compute definite integrals.
  • Verify the result you derived by hand or spot any errors in computation.
Having this technological aid boosts confidence in your solutions and serves as an excellent learning tool, especially for visual learners who benefit from seeing graphs of regions being revolved.

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Most popular questions from this chapter

Using Cross Sections Find the volumes of the solids whose bases are bounded by the graphs of \(y=x+1\) and \(y=x^{2}-1,\) with the indicated cross sections taken perpendicular to the \(x\) -axis. (a) Squares (b) Rectangles of height 1

The hydraulic cylinder on a woodsplitter has a 4 -inch bore (diameter) and a stroke of 2 feet. The hydraulic pump creates a maximum pressure of 2000 pounds per square inch. Therefore, the maximum force created by the cylinder is \(2000\left(\pi \cdot 2^{2}\right)=8000 \pi\) pounds. (a) Find the work done through one extension of the cylinder, given that the maximum force is required. (b) The force exerted in splitting a piece of wood is variable. Measurements of the force obtained in splitting a piece of wood are shown in the table. The variable \(x\) measures the extension of the cylinder in feet, and \(F\) is the force in pounds. Use the regression capabilities of a graphing utility to find a fourth-degree polynomial model for the data. Plot the data and graph the model. $$ \begin{array}{|l|l|c|c|c|c|c|c|} \hline x & 0 & \frac{1}{3} & \frac{2}{3} & 1 & \frac{4}{3} & \frac{5}{3} & 2 \\\ \hline F(x) & 0 & 20,000 & 22,000 & 15,000 & 10,000 & 5000 & 0 \\ \hline \end{array} $$ (c) Use the model in part (b) to approximate the extension of the cylinder when the force is maximum. (d) Use the model in part (b) to approximate the work done in splitting the piece of wood.

$$\pi \int_{0}^{2}\left[16-(2 y)^{2}\right] d y=2 \pi \int_{0}^{4} x\left(\frac{x}{2}\right) d x$$

Length of a Cable An electric cable is hung between two towers that are 40 meters apart (see figure). The cable takes the shape of a catenary whose equation is \(y=10\left(e^{x / 20}+e^{-x / 20}\right), \quad-20 \leq x \leq 20\) where \(x\) and \(y\) are measured in meters. Find the arc length of the cable between the two towers.

Volume of a Segment of a Paraboloid The region bounded by \(y=r^{2}-x^{2}, y=0\),and \(x=0\) is revolved about the the \(y\)-axis to form a paraboloid. A hole, centered along the axis of revolution, is drilled through this solid. The hole has a radius \(k\), 0<\(k\)<\(r\). Find the volume of the resulting ring (a) by integrating with respect to \(x\) and (b) by integrating with respect to \(y\).
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