Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

The base of a solid is bounded by one arch of \(y=\sqrt{\cos x},-\pi / 2 \leq x \leq \pi / 2\), and the \(x\) -axis. Each cross section perpendicular to the \(x\) -axis is a square sitting on this base. Find the volume of the solid.

Step-by-Step Solution

Answer

The volume of the solid is 2 cubic units.

Step 1: Understanding the Geometry

The base of the solid is the area under the curve given by the function \(y = \sqrt{\cos x}\) from \(x = -\pi/2\) to \(x = \pi/2\). The cross-sections perpendicular to the \(x\)-axis are squares, and their sides are determined by the length of the vertical segments from the \(x\)-axis to the curve \(y = \sqrt{\cos x}\).

Step 2: Area of Cross-Section

Since each cross-section is a square, the side length of the square at any position \(x\) is \(y = \sqrt{\cos x}\). Thus, the area of one square cross-section is \((\sqrt{\cos x})^2 = \cos x\).

Step 3: Set Up the Integral

To find the volume of the solid, integrate the area of the cross-sections along the \(x\)-axis from \(-\pi/2\) to \(\pi/2\). The volume \(V\) is given by the integral \(V = \int_{-\pi/2}^{\pi/2} \cos x \, dx\).

Step 4: Calculate the Integral

Compute the integral \(\int_{-\pi/2}^{\pi/2} \cos x \, dx\). The antiderivative of \(\cos x\) is \(\sin x\). Evaluate it from \(-\pi/2\) to \(\pi/2\) to get the volume: \([\sin x]_{-\pi/2}^{\pi/2} = \sin(\pi/2) - \sin(-\pi/2)\).

Step 5: Evaluate and Simplify

The value of \(\sin(\pi/2)\) is 1 and the value of \(\sin(-\pi/2)\) is -1. Substitute these values into the antiderivative result: \(1 - (-1) = 2\). Thus, the volume of the solid is 2.

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of calculus that focuses on finding integrals. It allows us to compute quantities like areas, volumes, and many other useful sums. When you think about an integral, imagine it as a sophisticated way to add up a continuous collection of numbers. In our context, integral calculus helps us find the total volume of a geometric shape, like the solid described in the exercise.
By setting up an integral, we're essentially summing up an infinite number of squares, each representing a small portion of the solid's volume. This is done through a definite integral, which provides a precise computation of the total volume based on the specified boundaries.

Solid Geometry

Solid geometry is the branch of geometry that deals with three-dimensional figures. Unlike plane geometry, which focuses on flat shapes like circles and squares, solid geometry explores objects with depth.
In our exercise, the solid has a base defined by the curve of the function \(y = \sqrt{\cos x}\) from \(x = -\pi/2\) to \(x = \pi/2\). Each slice or cross-section of this solid perpendicular to the \(x\)-axis forms a square. This 3D aspect of the problem is what makes it a solid geometry problem.

• It involves understanding the spatial relationship between curves and surfaces.
• Recognizing the way these elements interact to form a solid shape.

The volume here is determined by summing up the areas of these cross-sectional squares along the \(x\)-axis.

Volume Calculation

Calculating the volume of a solid like this requires integrating the area of its cross sections. Each cross-section is a square with side length \(y = \sqrt{\cos x}\). This side length becomes the base for our area calculation, making the area \((\sqrt{\cos x})^2 = \cos x\).
To find the entire volume, these areas are integrated from \(x = -\pi/2\) to \(x = \pi/2\). The process works like this:

• First, find the expression for the area of a single cross-section (a square) as a function of \(x\).
• Next, integrate this function across the specified interval.
• The result of this definite integral gives the volume of the solid.

The key here is understanding that the integral acts as a sum of all these infinitesimal areas, consumed together to form the total volume.

Definite Integral

A definite integral gives us a precise value representing the accumulation of quantities over an interval. It is expressed with two bounds indicating the start and end of this interval, and it is particularly useful when dealing with problems involving areas and volumes.
The definite integral of \(\int_{-\pi/2}^{\pi/2} \cos x \, dx\) in our exercise is used to calculate the total volume of the solid. Here’s how it works:

• First, identify the function to be integrated. In this case, it’s the area function \(\cos x\) representing the cross-section.
• Next, determine the bounds \(-\pi/2\) to \(\pi/2\), which are the limits of the base of the solid along the \(x\)-axis.
• Finally, compute the integral by finding the antiderivative and evaluating it at these bounds.

This yields the total volume, confirming the power of definite integrals in solving real-world geometry problems.