Problem 56 - Calculus Early Transcendentals

Question

Find the volume of the following solids using the method of your choice. The solid formed when the region bounded by \(y=\sin x\) and \(y=1-\sin x\) between \(x=\pi / 6\) and \(x=5 \pi / 6\) is revolved about the \(x\) -axis

Step-by-Step Solution

Answer

The volume of the solid is \(V = \pi(2\pi/3 - \sqrt{3})\).

Step 1: Graph the functions: \(y=\sin x\) and \(y=1-\sin x\)

In order to find the region, first, graph the functions \(y=\sin x\) and \(y=1-\sin x\). Identify the intersection points, which occur at \(x=\pi / 6\) and \(x=5\pi / 6\). These points will be the limits of integration. **Step 2: Set up the integral to find volume**

Step 2: Set up the disk integration

To find the volume of the solid when the region is revolved about the \(x\)-axis, use the disk method. For the disk method, the area of each disk is given by \(A(x)=\pi [(1-\sin x)^2 - (\sin x)^2]\). The volume of the solid is found by integrating the area of the disks along the \(x\)-axis from \(x=\pi / 6\) to \(x=5\pi / 6\). This gives the integral: $$ V =\int_{\pi / 6}^{5\pi / 6} \pi[(1-\sin x)^2 - (\sin x)^2] dx $$ **Step 3: Integrate and simplify**

Step 3: Integrate the integral for the volume

To solve the integral, first, we need to simplify the integrand: $$ (1-\sin x)^2 - (\sin x)^2 = 1 - 2 \sin x + \sin^2 x - \sin^2 x = 1 - 2 \sin x $$ Now, integrate the simplified integrand: $$ V =\int_{\pi / 6}^{5\pi / 6} \pi(1 - 2\sin x) dx = \pi \int_{\pi / 6}^{5\pi / 6} (1 - 2\sin x) dx $$ $$ V= \pi \left[x-2\int_{\pi / 6}^{5\pi / 6}\sin x dx \right]_{\pi / 6}^{5\pi / 6} $$ **Step 4: Evaluate the integral and find the volume**

Step 4: Evaluate the definite integral

Now evaluate the definite integral: $$ V= \pi \left[(5\pi / 6 - 2\int_{\pi / 6}^{5\pi / 6}\sin x dx)-(\pi/6 - 2\int_{\pi / 6}^{\pi/6}\sin x dx) \right] $$ The integral of \(\sin x\) is \(-\cos x\), so the initial value part: $$ 2\int_{\pi / 6}^{5\pi / 6}\sin x dx = 2[-\cos(5\pi / 6) +\cos(\pi / 6)] = \sqrt{3} $$ Plug the result back to the expression. $$ V= \pi \left[5\pi / 6 - \sqrt{3} - \pi/6\right] = \pi (2\pi/3 - \sqrt{3}) $$ Therefore, the volume of the solid formed when revolving the region bounded by \(y=\sin x\) and \(y=1-\sin x\) between \(x=\pi / 6\) and \(x=5 \pi / 6\) about the \(x\)-axis is: \(V = \pi(2\pi/3 - \sqrt{3})\).

Key Concepts

Volume of Solids of Revolution

The volume of solids of revolution refers to the volume that is formed when a region in the plane is rotated around a particular axis. One of the most popular approaches to solving such problems is using the Disk Method.
When a two-dimensional area is revolved around an axis, the resultant shape is a three-dimensional solid. To find the volume of this solid, we picture the region being sliced into thin perpendicular disks. Each disk has a small thickness and the volume of such a disk can be found by treating it as a cylinder with a calculable circular cross-section area.
The volume of the entire solid is then the summation of the volumes of these small disks. Think of stacking many coin-like cylinders to form an entire solid. The method allows for an elegant calculation using integration.
To apply the Disk Method:

• Determine the function that represents the outer or top curve (if revolved around the x-axis).
• Calculate the radius of each disk as the distance from the axis of rotation to the function line.
• Use the formula for the volume of a disk: \[ V = \pi \int_{a}^{b} R(x)^2\, dx \]
• Evaluate the definite integral over the specified interval.

Definite Integral Evaluation

Evaluating definite integrals is an essential part of finding the volume using Disk or Shell methods. A definite integral calculates the accumulation of quantities—such as areas or volumes—between specified limits.
When we integrate, we find the net area under a curve between two points, which also helps determine the quantity like volume when multiplied by an appropriate factor such as \(\pi\) for disk areas.
To evaluate a definite integral, follow these steps:

• Identify the appropriate integral that reflects the area or accumulated quantity of interest.
• Solve the indefinite integral of the function (antiderivative).
• Apply the fundamental theorem of calculus, which involves evaluating the antiderivative at the upper limit and subtracting it from the evaluation at the lower limit.
• Subtract the resulting numbers to find the total accumulated value between the limits.

For instance, in the given problem, integrating the expression \(1 - 2 \sin x\) involves handling the areas presented by the functions, giving us insight into how these areas contribute to the total volume of the solid.

Graphing Trigonometric Functions

Graphing trigonometric functions is crucial to understanding the visual representation of a problem and to identify bounds between which these are integrated.
Functions such as \(y = \sin x\) and \(y = 1 - \sin x\) create intriguing patterns and often intersect at key points. These intersections are important because they form the bounds for the integration process.
To graph trigonometric functions:

• Identify significant points where these functions cross or intersect each other.
• Understand their periodicity and amplitude, which determines how high or low the graph stretches along the y-axis.
• Use graphing software or sketching methods to visualize for accuracy.

These graphical insights are not just useful for setting integration limits but also simplify the understanding of spatial relationships in geometric interpretations.

Integration Techniques

Various integration techniques can simplify the process of integrating complex functions to find volumes or areas. Each technique is suitable depending on the form and characteristics of the function.
Some of these techniques include:

• Substitution Method: Suitable for functions that are products of an inner function and its derivative. Simplifies the integral into a more recognizable form.
• Integration by Parts: Useful for functions that are products of algebraic and transcendental terms. Follows the formula \( \int udv = uv - \int vdu \).
• Trigonometric Integrals: Often appear when integrating trigonometric functions, using identities to simplify the integrands.

With these techniques, integrals that appear complicated at first might become manageable and easier to evaluate, as seen in the problem where simplifying \((1 - \sin x)^2 - (\sin x)^2\) reduces the complexity of the integral.