Problem 33 - Thomas Calculus

Question

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 \(x=\ln 2 .\)

Step-by-Step Solution

Answer

The volume is \(2\pi (1 - \ln 2)\).

Step 1: 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 \( x = \ln 2 \).

Step 2: 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 \( y = e^{x} \).

Step 3: 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 \( b = \ln 2 \).

Step 4: 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 \( v = e^{x} \).

Step 5: Integrate by Parts

Apply the integration by parts formula: \[\int u \, dv = uv - \int v \, du.\]Here, it becomes: \[V = 2\pi \left[ (\ln 2 - x)e^{x} \bigg|_{0}^{\ln 2} + \int e^{x} \, dx \bigg|_{0}^{\ln 2} \right].\]

Step 6: 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 \(0 - \ln 2 = -\ln 2.\)

Step 7: 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 \(V = 2\pi (-\ln 2 + 1) = 2\pi(1 - \ln 2)\).

Step 8: Express the Final Result

The volume of the solid of revolution is given by multiplying the final expression by \(2\pi\): \(V = 2\pi (1 - \ln 2)\).

Key Concepts

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 \( e^x \).

This method is often convenient when the region is described easier in terms of \( x \) rather than \( y \). Remember to properly set the limits and express the radius and height correctly within the integral.

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 \( V = 2\pi \left[ (\ln 2 - x)e^x \bigg|_{0}^{\ln 2} + \int e^x \, dx \bigg|_{0}^{\ln 2} \right] \).

This method is particularly useful when integrals of certain parts are known or simpler, which helps simplify complex products of functions, as seen with the exponential term here. The choice of \( u \) and \( dv \) significantly affects the simplicity of solving these integrals.

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.

Understanding the geometric interpretation of each integral term helps visualize and compute the volume more efficiently. This enhances comprehension of how each element within the integral contributes to the total volume.

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.

These calculus concepts, especially when applied to physical and geometric problems, not only show how interconnected mathematical ideas can solve complex problems but also provide a means to model and understand various aspects of the physical world.