Problem 4 - Calculus Single Variable

Question

In each of Exercises 1-6, use the method of disks to calculate the volume \(V\) of the solid that is obtained by rotating the given planar region \(\mathcal{R}\) about the \(x\) -axis. \(\mathcal{R}\) is the region above the \(x\) -axis, below the graph of \(y=\) \(\sec (x),\) to the right of \(x=0,\) and to the left of \(x=\pi / 4\)

Step-by-Step Solution

Answer

The volume of the solid is \(\pi\).

Step 1: Understanding the Problem

We need to find the volume of a solid formed by rotating the region above the x-axis and below the curve \(y = \sec(x)\), bounded between \(x = 0\) and \(x = \frac{\pi}{4}\), about the x-axis.

Step 2: Setup of the Disk Method

The volume of the solid of revolution can be calculated using the disk method. The formula is given by \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]In this problem, \(f(x) = \sec(x)\), with limits of integration from \(x = 0\) to \(x = \frac{\pi}{4}\).

Step 3: Formulate the Integral

Substitute \(f(x) = \sec(x)\) into the disk method formula: \[ V = \pi \int_{0}^{\frac{\pi}{4}} (\sec(x))^2 \, dx \].

Step 4: Evaluate the Integral of \(\sec^2(x)\)

Recall that the integral of \(\sec^2(x)\) is \(\tan(x)\). Therefore, compute the integral:\[ \int_{0}^{\frac{\pi}{4}} \sec^2(x) \, dx = \left[ \tan(x) \right]_{0}^{\frac{\pi}{4}} \].

Step 5: Calculate the Definite Integral

Evaluate the antiderivative from \(x = 0\) to \(x = \frac{\pi}{4}\):\[ \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1 \].

Step 6: Compute the Volume

Substitute the result of the definite integral back into the volume formula:\[ V = \pi (1) = \pi \]. So, the volume of the solid is \(\pi\).

Key Concepts

Volume of Solids of Revolution

The idea of finding the volume of solids of revolution is like creating a 3D object by spinning a 2D shape around an axis. In this exercise, the 2D shape is a part of the graph defined by the function \(y = \sec(x)\). By rotating this shape about the \(x\)-axis, we create a 3D solid.
The disk method is often used to find such volumes. Imagine slicing the solid into thin circular disks perpendicular to the axis of rotation.

• Each disk's thickness is a very small change in \(x\), denoted as \(dx\).
• The radius of each disk is determined by the value of the function at that point, \(f(x) = \sec(x)\).
• The volume of one such disk is \(\pi \times [\text{radius}]^2 \times \text{thickness}\), or \(\pi [f(x)]^2 dx\).

Summing up the volumes of these disks from \(x = 0\) to \(x = \frac{\pi}{4}\) gives the total volume of the solid, leading us to the integral formula.
For this problem, the formula becomes \(V = \pi \int_{0}^{\frac{\pi}{4}} (\sec(x))^2 \, dx\), which we solve using integral calculus.

Integral Calculus

Integral calculus is like reverse engineering a function to learn about its overall behavior. It's used here to accumulate the infinitely small pieces, which in the case of the disk method, are the volumes of the disks.
The fundamental idea in integral calculus is finding antiderivatives, or functions whose derivative produces a given function.
Given the formula, \(V = \pi \int_{0}^{\frac{\pi}{4}} (\sec(x))^2 \, dx\), the task is to find the total area under the curve \(\sec^2(x)\) from \(x = 0\) to \(x = \frac{\pi}{4}\).
To do this, we need the antiderivative of \(\sec^2(x)\), which is \(\tan(x)\).
By evaluating this antiderivative at the upper and lower limits and finding the difference, we determine the overall contribution to the volume. In this specific problem, it simplifies nicely to give \(\tan\left(\frac{\pi}{4}\right) - \tan(0) = 1\).
Consequently, the calculated volume of the solid is \(\pi\), indicating that integral calculus offers a powerful approach to solving geometrically complex problems.

Trigonometric Integration

Trigonometric integration is a specialized area of calculus dealing with integrals of trigonometric functions. These functions often appear in areas involving circular or periodic phenomena.
One common integral we encounter is \(\int \sec^2(x)\,dx\), which directly provides \(\tan(x)\) as its antiderivative. Here’s why it’s useful:

• Trigonometric identities simplify expressions and make finding integrals manageable.
• For instance, knowing that \(\sec^2(x) = 1 + \tan^2(x)\) provides insights into how to integrate or manipulate expressions.
• These identities often play a crucial role when decomposing complex trigonometric expressions within integrals.

Understanding the antiderivative of \(\sec^2(x)\) helps us solve the given integral, leading to a precise calculation of the volume of the solid.
Trigonometric integration, therefore, is vital when dealing with functions that describe waves, circles, and rotations, allowing us to transform and solve such functions effectively.