Problem 70 - Essential Calculus

Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(x\) -axis. $$ y=\cos \frac{x}{2}, \quad y=\sin \frac{x}{2}, \quad x=0, \quad x=\pi / 2 $$

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by the curves about the \(x\)-axis is \( \pi^2/4 \)

Step 1: Establish the Bounded Region

Firstly, we need to plot the curves \(y = \sin(x/2)\) and \(y = \cos(x/2)\), along with the lines \(x=0\) and \(x = \pi/2\). When we sketch these, we see that the bounded region lies above the x -axis, meaning that we can use the disc method for rotating about the x-axis.

Step 2: Set Up the Integral for the Disk Method

The disk method formula for volume is \(V = \pi\int_{a}^{b}(g(x))^2 dx\), where [a,b] is the interval [0 , \(\pi/2\)] on the x-axis, and g(x) is the radius of the disk, which is here given by the upper function minus the lower function, i.e., \(g(x) = \cos(x/2) - \sin(x/2)\). Hence, set up this integral: \(V = \pi\int_{0}^{\pi/2}(\cos^2(x/2) - 2\cos(x/2)\sin(x/2) + \sin^2(x/2)) dx\).

Step 3: Simplify and Compute the Integral

The integrand simplifies to \(V = \pi\int_{0}^{\pi/2}(1 - \sin(x)) dx\), because the sum of squares of sine and cosine is one and the double-angle identity for sine can be expressed as \(\sin(2u) = 2\sin(u)*\cos(u)\). Then, compute the integral \(V = \pi [ x - \frac{1}{2}(-\cos(x)) ]_{0}^{\pi/2}\).

Step 4: Evaluate the Integral

Finally, evaluate the integral at the bounds \(x=0\) and \(x= \pi/2\). This yields \(V = \pi * \pi/4 = \pi^2/4\)

Key Concepts

Disk Method

The Disk Method is a technique used in calculus to find the volume of a solid of revolution. This method is particularly useful when a region is revolved around an axis, here, the x-axis in our example.
Imagine a thin slice, or disk, cut perpendicular to the axis of rotation from the solid. The volume of this disk can be calculated as the area of a circle (since the disk is circular) multiplied by its thickness.
- For a curve revolved around the x-axis, the formula for volume using the Disk Method is:
\[ V = \pi \int_{a}^{b} [g(x)]^2 \, dx \]Where:

• \(V\) is the volume of the solid.
• \(g(x)\) represents the radius of the disk, which can be expressed as the difference between the outer and inner functions, here \(\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)\).
• \([a,b]\) are the limits along the x-axis across which the region is rotated, \([0, \pi/2]\) in this exercise.

Understanding this method makes it easier to visualize how a flat shape can be transformed into a three-dimensional object.

Integral Calculus

Integral calculus helps us solve problems related to area, volume, and lengths of curves, among others. It is the mathematical tool we use to calculate the volume of the solid of revolution using definite integrals.
The basic idea is to sum an infinite number of infinitesimally small quantities (disks, in our case) to find a total quantity, which in this case is the volume.- The integral representations consist of three parts:

• Function to be integrated \((g(x))^2\), where our particular functions are \(\cos^2\left(\frac{x}{2}\right)-2\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)+\sin^2\left(\frac{x}{2}\right)\).
• Integration limits, which are from \(x=0\) to \(x=\pi/2\).
• The integration itself, where we use calculus rules and identities to simplify and solve, resulting in the final value of the volume.

By carrying out the integral calculation correctly, you determine the area under the curve, or here, the volume of rotated spaces, resulting in \(V = \frac{\pi^2}{4}\).

Trigonometric Functions

Trigonometric functions like sine and cosine play a vital role in integral calculus, especially in problems related to solids of revolution involving circular or oscillatory figures.
In this exercise, the solid's boundary is defined by the functions \(y=\cos\left(\frac{x}{2}\right)\) and \(y=\sin\left(\frac{x}{2}\right)\).
- These functions help establish the specific shape and area of the segment being rotated to create the solid.

• \(\cos\left(\frac{x}{2}\right)\) represents the upper boundary.
• \(\sin\left(\frac{x}{2}\right)\) is the lower boundary.
• The identity \(\cos^2(u) + \sin^2(u) = 1\) simplifies the integrand, crucial in obtaining a simple expression to compute the integral.

Understanding these trigonometric functions and identities helps in simplifying the integral and accurately calculating the solid's volume after rotation.