Problem 13 - Calculus: Early Transcendentals

Question

The base of a solid is the region between \(f(x)=\cos x\) and \(g(x)=-\cos x,-\pi / 2 \leq x \leq\) \(\pi / 2\), and its cross-sections perpendicular to the \(x\) -axis are squares. Find the volume of the solid.

Step-by-Step Solution

Answer

The volume of the solid is \( 2\pi \).

Step 1: Determine the Side Length of the Squares

The side length of each cross-sectional square is determined by the length between the functions \( f(x) = \cos x \) and \( g(x) = -\cos x \). Thus, the side length is given by the expression \( \text{Length} = \cos x - (-\cos x) = 2\cos x \).

Step 2: Establish the Cross-sectional Area

The area of a cross-section taken perpendicular to the \( x \)-axis is a square. Therefore, for each cross-section, the area \( A(x) \) is the square of the side length: \[ A(x) = (2\cos x)^2 = 4\cos^2 x. \]

Step 3: Integrate Area over the Given Interval

To find the total volume of the solid, integrate the area \( A(x) \) across the interval \(-\pi/2 \leq x \leq \pi/2\). The volume \( V \) is: \[ V = \int_{-\pi/2}^{\pi/2} 4\cos^2 x\, dx. \]

Step 4: Simplify and Solve the Integral

Use the trigonometric identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \) to rewrite the integral: \[ V = \int_{-\pi/2}^{\pi/2} 4\left( \frac{1 + \cos 2x}{2} \right)\, dx = 2\int_{-\pi/2}^{\pi/2} (1 + \cos 2x)\, dx. \] Split the integral: \[ V = 2\left( \int_{-\pi/2}^{\pi/2} 1\, dx + \int_{-\pi/2}^{\pi/2} \cos 2x\, dx \right). \]

Step 5: Evaluate the Integrals

The integral \( \int_{-\pi/2}^{\pi/2} 1\, dx = x \bigg|_{-\pi/2}^{\pi/2} = \pi \).For \( \int_{-\pi/2}^{\pi/2} \cos 2x\, dx \), use the substitution \( u = 2x \), then \( \frac{1}{2}\int_{-\pi}^{\pi} \cos u\, du = 0 \) because the cosine integral over a complete period is zero.

Step 6: Calculate Total Volume

Substituting back, we find \( V = 2\left( \pi + 0\right) = 2\pi. \) Therefore, the volume of the solid is \( 2\pi \).

Key Concepts

Integral Calculus

Integral Calculus is a powerful mathematical tool used to calculate areas, volumes, and integrals of functions. In this specific exercise, we are calculating the volume of a solid through integration. The essence of solving this problem involves understanding how a function represents the accumulation of smaller parts—in this case, squares.

To find the volume of the solid, we establish an integral over a specific interval. The key steps include:

• Determining an expression for the area of each cross-sectional slice.
• Setting up an integral that sums up these areas along the entire region of interest.
• Using integration techniques to solve the problem effectively.

This process leverages the fundamental theorem of calculus, which links differentiation and integration. It serves as the backbone for finding accumulated quantities, such as areas under curves or volumes of solids.

Cross-sectional Areas

A cross-sectional area is simply a slice of a three-dimensional object. This concept is crucial in calculating volumes of solids, especially in problems where the volume is not straightforward.

In our example, the solid's cross-section perpendicular to the x-axis is a square. The problem gives us two functions, \(f(x) = \cos x\) and \(g(x) = -\cos x\). The distance between these functions determines the side of each square:

• The distance is \(2\cos x\), as found by subtracting the lower function from the upper one.

This is the side of the square, and squaring it gives the area:

• Cross-sectional area: \(A(x) = (2\cos x)^2 = 4\cos^2 x\).

Applying the concept of cross-sectional areas allows for encapsulating complex structures into simpler planar shapes, easing the integration process.

Trigonometric Integration

Trigonometric Integration involves integrating expressions that contain trigonometric functions. This exercise utilizes the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\), which is a common trigonometric identity that simplifies the integration of \(\cos^2 x\).

The focus here is to make the integration process easier. Here's how it works in our problem:

• The volume is expressed as \(V = \int_{-\pi/2}^{\pi/2} 4\cos^2 x\, dx\).
• Simplifying using the identity transforms it into \(2 \int_{-\pi/2}^{\pi/2} (1 + \cos 2x)\, dx\).

Further, the integral splits into two components:

• A constant term which is straightforward to integrate.
• A trigonometric term where periodic properties simplify its evaluation, particularly over symmetric intervals.

With this method, you find yourself tackling problems that initially seem complex but break down beautifully with the right identities and techniques.