Problem 9 - Calculus Early Transcendentals

Question

Use the general slicing method to find the volume of the following solids. The solid whose base is the region bounded by the curve \(y=\sqrt{\cos x}\) and the \(x\) -axis and whose cross sections through the solid perpendicular to the \(x\) -axis are isosceles right triangles with a horizontal leg in the \(x y\) -plane and a vertical leg above the \(x\) -axis

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{1}{2}\).

Step 1: Find the area of a cross-sectional slice

The cross-section is an isosceles right triangle with a horizontal leg in the \(xy\)-plane and a vertical leg above the \(x\)-axis. Since the horizontal and vertical legs are equal, we can represent the area of the isosceles right triangle as: \(A(x) = \frac{1}{2}(\text{Horizontal Leg Length})^2 = \frac{1}{2}(y)^2 = \frac{1}{2}(\sqrt{\cos x})^2 = \frac{1}{2}\cos x\)

Step 2: Find the interval for x

Since the curve \(y = \sqrt{\cos x}\) lies within the interval \([0, \frac{\pi}{2}]\), we must integrate from \(0\) to \(\frac{\pi}{2}\).

Step 3: Set up the integral

We set up the integral to find the total volume of the solid by integrating the area function over the x-axis interval: \(V = \int_{0}^{\frac{\pi}{2}} A(x) dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{2}\cos x \, dx\)

Step 4: Evaluate the integral

Evaluate the integral to find the volume: \(V = \int_{0}^{\frac{\pi}{2}} \frac{1}{2}\cos x \, dx = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \cos x \, dx\) The integral of \(\cos x\) is \(\sin x\), so: \(V = \frac{1}{2}(\sin x\big|_{0}^{\frac{\pi}{2}}) = \frac{1}{2}(\sin \frac{\pi}{2} - \sin 0) = \frac{1}{2}(1 - 0) = \frac{1}{2}\) Therefore, the volume of the solid is \(\frac{1}{2}\).

Key Concepts

General Slicing Method

The General Slicing Method is a powerful technique used in calculus to determine the volume of a solid. It's especially useful for solids with irregular shapes. To apply this method, imagine slicing the solid into thin, parallel pieces, often perpendicular to the base of the solid. Each slice is a cross-section, and the basic idea is to sum up the volumes of these slices to find the total volume of the solid.

Here's a step-by-step breakdown of the method:

• **Identify the cross-sectional shape:** For the given problem, cross-sections are isosceles right triangles.
• **Determine the formula for the area of a slice:** This involves identifying the dimensions of the slice that change as you move along the axis of integration.
• **Integrate the area formula over the desired interval:** This step computes the total volume by integrating the area of each slice over the range it spans on the axis.

General Slicing is a flexible method that can be adjusted according to the cross-sectional shape and the region of the solid's base. Remember, each slice's volume is its area multiplied by its thickness (infinitesimally small). Given a function that defines the curve of the base, it becomes possible to utilize definite integrals to find the volume.

Isosceles Right Triangles

Isosceles right triangles have a distinct shape which helps in simplifying volume calculations. In these triangles, two legs are equal in length, and the angles are 45°, 45°, and 90°. For our problem, the horizontal and vertical legs are of equal length, and these triangles are used as cross-sections.

The unique properties of isosceles right triangles allow us to easily calculate their area. Since the legs are equal, the formula for their area is:

• \[ A = \frac{1}{2} \times (\text{leg length})^2 \]

This formula simplifies our work when we know one leg. In this problem, the leg is defined by the curve equation \( y = \sqrt{\cos x} \). This means the horizontal leg length also equals \( \sqrt{\cos x} \), which in turn influences both the triangle's area and the overall volume calculation.
Understanding how these triangles fit into the broader problem allows us to correctly set up the integral needed for the volume calculation.

Definite Integrals

Definite integrals are central to calculus and are crucial when calculating the volume of solids. A definite integral helps find the accumulated sum or total when summing infinitely small pieces across a defined interval. In the context of volume calculation, it adds up the tiny volumes of the slices modeled by a single function.

In the given exercise, once the area function \( A(x) \) is established based on the geometry of the cross-section, we use definite integrals to sum these areas over the interval of interest. Here, the interval is \([0, \frac{\pi}{2}]\), where the function related to our slices is integrated:

• \[ V = \int_{0}^{\frac{\pi}{2}} \frac{1}{2} \cos x \, dx \]

When we evaluate a definite integral, it involves finding the antiderivative, followed by computing the difference in its values at the bounds of the interval. For this problem, the antiderivative of \( \cos x \) is \( \sin x \), leading to the final calculation of the volume as \( \frac{1}{2} \).
Understanding definite integrals this way opens up the ability to tackle diverse problems involving volume, regardless of the specifics of the shape or function involved.