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Magazine Chapter 8: Problem 33
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 \(x=\ln 2\) about the line \(x=\ln 2 .\)
Short Answer
Expert verified
The volume is \(2\pi (1 - \ln 2)\).
Step by step solution
01
Understand the Problem
We need to find the volume of a solid of revolution. The region of interest is bounded by the curve \( y = e^{x} \), the \( x \)-axis, and the line \( x = \ln 2 \). We will use the method of shells to revolve this region around the line \( x = \ln 2 \).
02
Set Up the Shell Method Formula
The shell method formula for revolving around a vertical line is given by \( 2\pi \int_{a}^{b} (radius)(height)\,dx \). The radius in this context is the distance from a point \( x \) to the line \( x = \ln 2 \), which is \( (\ln 2 - x) \). The height is given by \( y = e^{x} \).
03
Determine the Limits of Integration
The limits of integration define the range over which we'll compute the volume. The region is bounded by \( x = 0 \) and \( x = \ln 2 \) along the \( x \)-axis, making these the limits of integration: \( a = 0 \) and \( b = \ln 2 \).
04
Set Up and Evaluate the Integral
Using the shell method, the integral becomes: \[V = 2\pi \int_{0}^{\ln 2} (\ln 2 - x)e^{x} \,dx.\]We need to integrate by parts to solve this integral. Let \( u = \ln 2 - x \) and \( dv = e^{x} \, dx \); then \( du = -1 \, dx \) and \( v = e^{x} \).
05
Integrate by Parts
Apply the integration by parts formula: \[\int u \, dv = uv - \int v \, du.\]Here, it becomes: \[V = 2\pi \left[ (\ln 2 - x)e^{x} \bigg|_{0}^{\ln 2} + \int e^{x} \, dx \bigg|_{0}^{\ln 2} \right].\]
06
Evaluate the Integrated Parts
Calculate \((\ln 2 - x)e^{x} \bigg|_{0}^{\ln 2}\): At \(x = \ln 2\), \((\ln 2 - \ln 2)e^{\ln 2} = 0\). At \(x = 0\), \((\ln 2 - 0)e^{0} = \ln 2\). Thus, this part is \(0 - \ln 2 = -\ln 2.\)
07
Complete the Integration Process
Continue the evaluation process: \[\int e^{x} \, dx \bigg|_{0}^{\ln 2} = \left.e^{x} \right|_{0}^{\ln 2} = e^{\ln 2} - e^{0} = 2 - 1 = 1.\]. Thus the volume \(V = 2\pi (-\ln 2 + 1) = 2\pi(1 - \ln 2)\).
08
Express the Final Result
The volume of the solid of revolution is given by multiplying the final expression by \(2\pi\): \(V = 2\pi (1 - \ln 2)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
shell method
The shell method is a powerful technique in calculus used to find the volume of a solid of revolution. This method involves considering cylindrical "shells" that are formed when a region in the plane is revolved about an axis. For example, in this exercise, we revolve the region bounded by the exponential curve \( y = e^x \), the \( x \)-axis, and the line \( x = \ln 2 \) around the vertical line \( x = \ln 2 \).
- We calculate the volume by "summing up" the volume of these thin cylindrical shells.
- The shell method formula for revolving around a vertical line is \( V = 2\pi \int_{a}^{b} (radius)(height)\,dx \).
- If you are revolving around the vertical line \( x = k \), then the radius of a shell is \( |k - x| \). In our problem, it's \( \ln 2 - x \).
- The height is the function value \( f(x) \), here being \( e^x \).
integration by parts
Integration by parts is a technique that stems from the product rule for differentiation. It is used to integrate products of functions. In our example, we encountered the integral \( \int u \, dv = uv - \int v \, du \), which had to be solved using this method. Here's how it works in this context:
- Choose \( u \) and \( dv \) such that their derivatives and integrals respectively simplify the process. Here \( u = \ln 2 - x \) and \( dv = e^x \, dx \).
- Calculate \( du = -dx \) and \( v = e^x \).
- Substitute in to get \( V = 2\pi \left[ (\ln 2 - x)e^x \bigg|_{0}^{\ln 2} + \int e^x \, dx \bigg|_{0}^{\ln 2} \right] \).
volume of a solid
The volume of a solid of revolution refers to the three-dimensional space occupied by the solid that is created when a two-dimensional area is revolved around an axis. In this exercise, the area under the curve \( y = e^x \) from \( x = 0 \) to \( x = \ln 2 \), when revolved around \( x = \ln 2 \), creates the solid.
- The process of revolving a shape about an axis transforms it into a solid with thickness and volume.
- For accurate computation, calculus methods such as the disk method, washer method, or shell method can be applied based on the axis of rotation and symmetry.
- In our example, the shell method provided a convenient approach, considering the perpendicular distance of shells from the axis of rotation and the curve's height.
calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It's essential for understanding changes and areas under curves, which can describe real-world phenomena in physics, engineering, economics, etc. In this exercise:
- We used integrals, a key concept in calculus, to find the volume of the solid formed by revolving an area.
- The formula \( V = 2\pi \int_{a}^{b} (radius)(height)\, dx \) is applied using the fundamental theorem of calculus for definite integrals.
- Integration by parts, another technique within calculus, helped simplify the integral involving exponential functions, thus solving the volume problem.
