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Magazine Chapter 9: Problem 35
A solid has as its base the region bounded by the curves \(y=e^{x}\) and \(y=e^{-x}\) and the line \(x=1\). If every plane section perpendicular to the \(x\) axis is a square find the volume of the solid.
Short Answer
Expert verified
The volume of the solid is \(e^{2} - e^{-2} - 4\).
Step by step solution
01
- Identify the bounds
Analyze the given bounds of the solid. The base is the region bounded by the curves \(y = e^{x}\) and \(y = e^{-x}\) up to the line \(x = 1\). This means we need to consider the interval for \(x\) from \( -1 \) to \(1\), since \(e^{-x}\) and \(e^{x}\) intersect at \(x = 0\).
02
- Area of the cross section
Each cross-section perpendicular to the \(x\)-axis is a square. The side length of the square at position \(x\) is the distance between the curves \(y = e^{x}\) and \(y = e^{-x}\). Therefore, the side length is \(e^{x} - e^{-x}\). The area of a square is the side length squared. So, the area \(A(x)\) of each cross-section is given by \((e^{x} - e^{-x})^{2}\).
03
- Set up the integral for volume
To find the volume of the solid, integrate the area of the cross-sections along the \(x\)-axis from \(x = -1\) to \(x = 1\): \(\text{Volume} = \int_{-1}^{1} (e^{x} - e^{-x})^{2} \, dx \).
04
- Simplify the integrand
Simplify \((e^{x} - e^{-x})^{2}\) using the identity \((a - b)^{2} = a^{2} - 2ab + b^{2}\): \((e^{x} - e^{-x})^{2} = e^{2x} - 2 \cdot e^{x} \cdot e^{-x} + e^{-2x} = e^{2x} - 2 + e^{-2x}\).
05
- Integrate
Now, integrate the simplified expression: \(\text{Volume} = \int_{-1}^{1} (e^{2x} - 2 + e^{-2x}) \, dx\). Split the integral into three separate integrals: \(\text{Volume} = \int_{-1}^{1} e^{2x} \, dx - \int_{-1}^{1} 2 \, dx + \int_{-1}^{1} e^{-2x} \, dx\).
06
- Compute the integrals
Evaluate each integral separately: 1. \(\int_{-1}^{1} e^{2x} \, dx = \[ \frac{e^{2x}}{2} \]_{-1}^{1} = \frac{e^{2} - e^{-2}}{2}\). 2. \(\int_{-1}^{1} 2 \, dx = 2 \[ x \]_{-1}^{1} = 2 \cdot (1 - (-1)) = 4\). 3. \(\int_{-1}^{1} e^{-2x} \, dx = \[ -\frac{e^{-2x}}{2} \]_{-1}^{1} = -\frac{e^{-2}}{2} + \frac{e^{2}}{2} = \frac{e^{2} - e^{-2}}{2}\).
07
- Combine the results
Combine the results from the integrals: \(\text{Volume} = \frac{e^{2} - e^{-2}}{2} - 4 + \frac{e^{2} - e^{-2}}{2} = e^{2} - e^{-2} - 4\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral Calculus is a major branch of calculus that deals with integrals and their properties. Integrals are used to find accumulations such as areas under a curve, volumes of solids, and other quantities that sum up continuously changing values. In this exercise, we use integral calculus to determine the volume of a solid of revolution. This involves setting up an integral that represents the volume based on the cross-sectional area of the solid.
- The key idea is to integrate a function over a specific interval to find the total accumulation.
- In our exercise, the cross-sectional area is squared and then integrated along the bounds of the given region.
Definite Integrals
Definite Integrals are a type of integral that have specific limits or bounds. They are used to calculate the accumulated value of a function from one point to another. In the context of our exercise, we use definite integrals to find the volume of the solid bounded by the given curves.
\[ \text{Volume} = \int_{-1}^{1} (e^{x} - e^{-x})^{2} \, dx \]
This integral is then split and evaluated to find the final volume.
- The definite integral provides the exact total area under a curve between two points, which is crucial for calculating volumes.
- It combines the integrand (the function being integrated) with the limits of integration to give a precise result.
\[ \text{Volume} = \int_{-1}^{1} (e^{x} - e^{-x})^{2} \, dx \]
This integral is then split and evaluated to find the final volume.
Cross-sectional Area
The Cross-sectional Area is the area of a shape that is obtained by cutting through a solid perpendicularly to a specific axis. For the exercise, we are particularly interested in the cross-sections perpendicular to the \(x\)-axis.
- Each cross-section is a square whose side length comes from the distance between the curves \(y = e^{x}\) and \(y = e^{-x}\).
- The side length at any position \(x\) is \(e^{x} - e^{-x}\), and since the area of a square is its side length squared, the area function becomes \((e^{x} - e^{-x})^{2}\).
- First, find the side length: \(e^{x} - e^{-x}\)
- Next, compute the area of the square: \( (e^{x} - e^{-x})^{2} \)
- Finally, set up and evaluate the integral of the area function over the interval from \(-1\) to \(1\).
