Problem 79 - Thomas Calculus

Question

Volume A region in the first quadrant is bounded above by the curve \(y=\cosh x,\) below by the curve \(y=\sinh x,\) and on the left and right by the \(y\) -axis and the line \(x=2,\) respectively. Find the volume of the solid generated by revolving the region about the \(x\) -axis.

Step-by-Step Solution

Answer

The volume is \(2\pi\).

Step 1: Visualizing the Region

First, we identify the region we will revolve. The curves are given by \(y = \cosh x\) and \(y = \sinh x\), bounded on the left by the \(y\)-axis (which is \(x = 0\)) and on the right by the line \(x = 2\). This creates a region between these two hyperbolic functions from \(x = 0\) to \(x = 2\).

Step 2: Setting Up the Volume Integral

The volume of the solid generated by revolving the region around the \(x\)-axis can be calculated using the disk method. The relevant formula is: \[ V = \pi \int_{a}^{b} (R^2 - r^2) \, dx \] where \(R = \cosh x\) is the outer radius and \(r = \sinh x\) is the inner radius.

Step 3: Simplifying the Integral

Substitute \(R = \cosh x\) and \(r = \sinh x\) into the formula: \[ V = \pi \int_{0}^{2} ((\cosh x)^2 - (\sinh x)^2) \, dx \] Using the identity \((\cosh x)^2 - (\sinh x)^2 = 1\), the integral simplifies to: \[ V = \pi \int_{0}^{2} 1 \, dx \]

Step 4: Calculating the Definite Integral

Evaluate the integral \(\int_{0}^{2} 1 \, dx \). This is a straightforward integral where: \[ \int_{0}^{2} 1 \, dx = [x]_{0}^{2} = 2 - 0 = 2 \]

Step 5: Final Volume Calculation

Plug the result back into the formula for volume: \[ V = \pi \times 2 = 2 \pi \] Thus, the volume of the solid generated by revolving the region about the \(x\)-axis is \(2\pi\).

Key Concepts

Disk Method

The Disk Method is a powerful technique used to find the volume of a solid of revolution. When an area or region is revolved around an axis (most often the x-axis or y-axis), it forms a 3D solid. The Disk Method involves slicing this solid vertically (perpendicular to the axis of rotation) into thin disks.

• Each disk has a thickness "dx" when the solid is revolved around the x-axis.
• The radius of each disk comes from the function value at a specific point.

The volume of each individual disk is the area of a circle times its thickness, which is given by \[ \pi \cdot (\text{radius})^2 \cdot dx \]. When these thin disks are added from one boundary to another using integration, they give the total volume of the solid. The formula is \[ V = \pi \int_{a}^{b} (R^2 - r^2) \, dx \], where "R" is the radius from the outer curve and "r" is the radius from the inner curve. This helps account for the hollow part of a solid when both outer and inner curves are given.

Hyperbolic Functions

Hyperbolic functions may sound complex, but they are quite similar to their trigonometric cousins. While trigonometric functions relate to circles, hyperbolic functions relate to hyperbolas. The basic ones are the hyperbolic sine function, \( \sinh x \), and the hyperbolic cosine function, \( \cosh x \).

• \( \sinh x = \frac{e^x - e^{-x}}{2} \)
• \( \cosh x = \frac{e^x + e^{-x}}{2} \)

These functions are used to solve many mathematical problems, especially those involving hyperbolas or certain types of differential equations. An important identity in hyperbolic functions that aids in simplification is \[ (\cosh x)^2 - (\sinh x)^2 = 1 \].This identity is central in simplifying volume of revolution problems as it effectively reduces the complexity of the integral, as seen in the original problem.

Definite Integral Calculation

The definite integral is a core concept in calculus used to find areas, volumes, and the sum of infinite series, among other things. In this context, we're using the definite integral to find the volume of a solid.The setup involves integrating from a lower bound \(a\) to an upper bound \(b\). The function within the integral represents the height or area of the shapes being summed.For the original exercise, the definite integral was \[ \int_{0}^{2} 1 \, dx \], which calculates the area under the function over the interval from 0 to 2. This integral simplifies to evaluating \[ [x]_{0}^{2} \]. The evaluation involves subtracting the function value at the lower limit from the function value at the upper limit, resulting in \[ 2 - 0 = 2 \].This result is then used to find the volume, by multiplying by \( \pi \), to give the final volume of the solid as \( 2\pi \). The definite integral thus converts a geometric problem into an algebraic process manageable through calculus.