Question

Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=e^{-x}, y=e^{x}, x=0,\) and \(x=\ln 4\) revolved about the \(x\) -axis

Short answer

In the context of this problem, the volume of the solid generated by revolving the region bounded by the curves \(y=e^{-x}\), \(y=e^{x}\), \(x=0,\) and \(x=\ln 4\) around the \(x\)-axis is calculated using the cylindrical shells method. To do so, we set up and evaluate the following integral: \(V = \int_{0}^{\ln 4} 2\pi (e^{-x} - e^x) dx\) Solving this integral results in a volume of \(\frac{9\pi}{2}\) cubic units.

Step-by-step solution

Sketch the region of interest

To better visualize the problem, we need to sketch the region, including all of the given curves: \(y=e^{-x}\), \(y=e^{x}\), \(x=0,\), and \(x=\ln 4\). The region we are revolving around the \(x\)-axis is the area enclosed by these curves. Do this by making a rough sketch of the curves on the Cartesian plane or use a graphing calculator or software.

Set up the integral using the cylindrical shells method

To find the volume of the solid, we will use the cylindrical shells method, which involves finding the integral of the product of the circumference, height, and thickness of each cylindrical shell produced by the region when revolved around the \(x\)-axis. Consider a small vertical strip at a position \(x\) between \(x=0\) and \(x=\ln 4\), with a height of \(y\). When revolving around the \(x\)-axis, the strip will form a cylindrical shell with a radius of \(y\), height of \(\Delta x\), and circumference \(2\pi y\). The volume of this cylindrical shell is given by \(dV = 2\pi y \Delta x\). Since we are revolving the region around the \(x\)-axis, the height of this strip at any given \(x\) lies within the curves \(y=e^{-x}\) and \(y=e^x\). So the height of the strip can be calculated by finding the difference between these functions: \(y = e^{-x} - e^x\). Now set up the integral to find the volume: \(V = \int_{a}^{b} (2\pi y) \Delta x = \int_{0}^{\ln 4} 2\pi (e^{-x} - e^x) dx\)

Evaluate the integral and find the volume

Evaluate the integral to find the volume: \(V = \int_{0}^{\ln 4} 2\pi (e^{-x} - e^x) dx\) First, we use the linearity of integration and distribute the constant: \(V = 2\pi \int_{0}^{\ln 4} (e^{-x} - e^x)dx\) Now, we evaluate each part of the integral separately: \(V = 2\pi \left[ -e^{-x} - e^x \right]_{0}^{\ln 4}\) Substitute the upper and lower bounds into the expression: \(V = 2\pi \left[(-(1/4) - 4) - (-(1) - 1) \right]\) Simplify the expression: \(V = 2\pi \left[ -\frac{17}{4} + 2 \right]\) \(V = 2\pi \left[-\frac{9}{4}\right]\) Finally, compute the volume: \(V = -\dfrac{9\pi}{2}\) Since the volume must be a positive value, we take the absolute value: \(V = \dfrac{9\pi}{2}\) Therefore, the volume of the solid of revolution is \(\frac{9\pi}{2}\) cubic units.

Key concepts

Volume of Solid of Revolution

Finding the volume of a solid of revolution involves determining the volume of a 3D object created when a 2D region is revolved around an axis. This process is integral in calculus for understanding the characteristics of shapes formed by rotation. The area between curves can be bounded by various functions and lines, necessitating a visualization of these intersections before computation.

• In our problem, the region is bounded by the curves: \(y=e^{-x}\), \(y=e^x\), \(x=0\), and \(x=\ln 4\).
• When this region is rotated about the \(x\)-axis, it forms a solid whose volume we need to calculate.
• Sketching the region helps to conceptualize the boundaries and the shape of the solid formed after rotation.

Understanding the role of these boundaries allows us to accurately set up the integration process that will follow, leading to the calculation of the desired volume.

Cylindrical Shells Method

The Cylindrical Shells Method is an efficient way to find the volume of solids formed by rotating regions around an axis. This method is especially useful when dealing with rotations around the vertical axis or when horizontal slicing is complex.
Here’s how it works:

• Each small vertical strip of the original region acts as a cylindrical shell upon rotation.
• The height of the shell is given by the difference in values between the upper and lower curves (e.g., \(e^{-x} - e^x\)).
• The radius of the shell corresponds to the \(y\)-value of the strip, given by \(x\).
• The formula for the volume of one shell: \(dV = 2\pi \times\text{(radius)} \times\text{(height)} \times\Delta x\).

By integrating this expression over the interval of interest \([0, \ln 4]\), we find the total volume of the solid.

Definite Integrals

Definite integrals are crucial when calculating areas and volumes as they provide the means to summing infinitesimally small quantities over a continuous range. They are used to accumulate values of a function within specified limits, providing a total change or accumulation. In calculus, definite integrals help directly compute physical quantities like area under a curve or, as in our problem, the volume of a solid.
For the problem at hand:

• The integral expression \(\int_{0}^{\ln 4} 2\pi (e^{-x} - e^x) dx\) calculates the total volume by summing the volume contributions of all cylindrical shells between \(x=0\) and \(x=\ln 4\).
• Integrating \(e^{-x}\) and \(e^x\) separately simplifies the evaluation of the volume.
• The bounds of 0 and \(\ln 4\) ensure the computation reflects the entire region defined by the curves.

By accurately evaluating this integral, we compute the volume of the solid, which can sometimes lead to further simplification or the need for absolute values to ensure realistic, positive volume measures.

Exponential Functions

Exponential functions are essential in describing a broad range of natural phenomena, such as growth and decay processes. In calculus, these functions frequently appear owing to their unique properties and behavior, like constant relative growth rate and straightforward differentiation and integration.
Key features of these functions include:

• The basic form \(y = e^x\) where \(e\) is the base, representing an irrational constant approximately equal to 2.718.
• \(y = e^{-x}\) expresses rapid decay, decreasing as \(x\) increases.

For our exercise:

• These exponential functions form the upper and lower boundaries of the given region.
• The challenge is grasping the spread and intersection points of these functions, to visualize the bounded area accurately.

Understanding their properties and how they behave when manipulated through integration is vital in accurately calculating volumes or areas under curves they establish.