Problem 21 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises \(17-22\) about the \(x\) -axis. $$ y=\sqrt{\cos x}, \quad 0 \leq x \leq \pi / 2, \quad y=0, \quad x=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Boundary Function and Region

The given boundary function is \( y = \sqrt{\cos x} \) and the limits are \( 0 \leq x \leq \pi/2 \). We are revolving this area about the \( x\)-axis to find the volume of the solid.

Step 2: Use the Disk Method Formula

To find the volume of the solid of revolution, we use the disk method, which is given by:\[V = \int_{a}^{b} \pi [f(x)]^2 \,dx\]where \( f(x) = \sqrt{\cos x} \), \( a = 0 \), and \( b = \pi/2 \).

Step 3: Set Up the Integral

Substitute \( f(x) = \sqrt{\cos x} \) into the formula:\[V = \int_{0}^{\pi/2} \pi \left(\sqrt{\cos x}\right)^2 \, dx = \int_{0}^{\pi/2} \pi \cos x \, dx\]

Step 4: Evaluate the Integral

Compute the integral:\[V = \pi \int_{0}^{\pi/2} \cos x \, dx\]The antiderivative of \( \cos x \) is \( \sin x \). Therefore, evaluate the bounds:\[V = \pi [ \sin x ]_{0}^{\pi/2} = \pi (\sin(\pi/2) - \sin(0)) = \pi (1 - 0) = \pi\]

Step 5: Summarize the Result

The volume of the solid generated by revolving the region bounded by \( y = \sqrt{\cos x} \), \( y = 0 \), and \( x = 0 \) around the \( x\)-axis from \( x = 0 \) to \( x = \pi/2 \) is \( \pi \).

Key Concepts

Disk Method

When it comes to finding the volume of solids formed by revolving a region around an axis, the disk method is an efficient technique. It's particularly useful when the solid is generated by rotating a region bounded by a function around the x-axis. Here’s how it works:

• Visualize slicing the solid into thin, circular disks perpendicular to the x-axis.
• The thickness of each disk corresponds to a small change in the x-value, denoted as \( dx \).
• The radius of each disk is determined by the value of the function at that particular slice, \( f(x) \).

The disk method formula is \( V = \int_{a}^{b} \pi [f(x)]^2 \,dx \), where \( [f(x)]^2 \) gives the area of the circular face of each disk. By integrating from \( a \) to \( b \), you sum up all the infinitesimally thin disks to find the total volume.

Integral Calculus

Integral calculus is a fundamental part of calculus concerned with the concept of integration. In the context of volume of solids of revolution:

• An integral can be understood as the limit of a sum, which adds up an infinite number of infinitesimally small quantities.
• Integration helps to calculate not only areas under curves but also volumes of solids formed by those curves.
• In our scenario, integrating \( \pi \left(\sqrt{\cos x}\right)^2 \) over a specific interval allows us to determine the volume of the entire revolved region.

Integral calculus makes it possible to generalize solutions for a variety of problems involving curves and their properties over an interval. It effectively transforms geometric problems into calculations manageable by algebraic means.

Definite Integral Evaluation

The process of evaluating a definite integral involves solving an integral over a specified interval. In the exercise, we need to find the definite integral from 0 to \( \pi/2 \) of the function \( \pi \cos x \). This is the final step in applying the disk method.Here's a step-by-step breakdown:

• Identify the function to be integrated: \( \pi \cos x \).
• Find the antiderivative. The antiderivative of \( \cos x \) is \( \sin x \), so the antiderivative of \( \pi \cos x \) is \( \pi \sin x \).
• Evaluate this antiderivative at the two ends of the interval (0 and \( \pi/2 \)), and calculate the difference: \( \pi [ \sin(\pi/2) - \sin(0) ] \).

This results in a volume of \( \pi \), which exemplifies the elegance and utility of using definite integrals to solve real-world geometric problems.