Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the solid trapped between the surface \(z=\cos x \cos y\) and the \(x y\)-plane, where \(-\pi \leq x \leq \pi\), \(-\pi \leq y \leq \pi\).

Step-by-Step Solution

Answer

The volume of the solid is 0.

Step 1: Identify the limits of integration

The solid is bounded by the surface described by the function \(z = \cos x \cos y\) and the \(xy\)-plane. The given limits for \(x\) and \(y\) are \(-\pi \leq x \leq \pi\) and \(-\pi \leq y \leq \pi\). Thus, these will be the limits for our double integral to find the volume.

Step 2: Set up the double integral

The volume \(V\) of the solid beneath the surface \(z = \cos x \cos y\) and above the \(xy\)-plane can be found using a double integral: \[V = \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \cos x \cos y\, dy \, dx.\]

Step 3: Calculate the inner integral

First, calculate the inner integral with respect to \(y\): \[\int_{-\pi}^{\pi} \cos y \, dy.\] The integral of \(\cos y\) is \(\sin y\). Evaluating from \(-\pi\) to \(\pi\), we get:\[[\sin y]_{-\pi}^{\pi} = \sin(\pi) - \sin(-\pi) = 0 - 0 = 0.\]

Step 4: Evaluate the outer integral

Now substitute the result from the inner integral back into the outer integral:\[\int_{-\pi}^{\pi} 0 \cdot \cos x \, dx.\]Since the inner integral evaluated to zero, multiplying \(\cos x\) by zero results in the entire integral being zero.

Step 5: Conclude the calculation

Evaluate the outer integral, which results in:\[\int_{-\pi}^{\pi} 0 \, dx = 0.\]Thus, the volume of the solid is 0.

Key Concepts

Double Integrals

Double integrals are a powerful mathematical tool used to find volumes under a surface over a particular region. In this problem, we're asked to find the volume of a solid bounded by the surface described by the equation \( z = \cos x \cos y \). To solve this, we use double integrals, allowing us to integrate a function over two variables, typically \( x \) and \( y \).The process involves setting up an integral where we first integrate the function with respect to one variable and then use the result to integrate with respect to the second variable. This helps us capture the cumulative contribution of all infinitesimally small sections of the function over the defined region. It's like summing up endless slices of volume between the surface and the plane, inside the region bounded by the limits given for \( x \) and \( y \).For a double integral \( \int \int f(x, y) \, dy \, dx \), we perform the inner integral first, with respect to \( y \), and then use that result to evaluate the outer integral with respect to \( x \). This solidifies its utility in calculating the total volume enclosed.

Limits of Integration

When working with double integrals, understanding the limits of integration is crucial. Limits of integration define the boundaries within which the integration will take place.In the given problem, the function is \( z = \cos x \cos y \), and it is bounded by the cube \(-\pi \leq x \leq \pi\) and \(-\pi \leq y \leq \pi\). These intervals direct us on where to start and stop our integrals for \( x \) and \( y \).- **Inner Integral Limits:** These are often the limits of the second integral and dictate the range for integrating with respect to \( y \).- **Outer Integral Limits:** These are the initial bounds that apply to the variable \( x \) when our inner integral gets plugged right into the outer.Understanding these limits ensure that the entire volume is considered and different regions are not skipped during computation.

Cosine Function

The cosine function is an essential element in this exercise and contributes to defining the surface over which we integrate. For the function \( z = \cos x \cos y \), both variables \( x \) and \( y \) affect the height of the surface above the \( xy \)-plane.The cosine function oscillates between -1 and 1. When multiplying \( \cos x \) by \( \cos y \), the product will equally oscillate, influencing the volume calculation. Specifically, both \( \sin(\pi) \) and \( \sin(-\pi) \) equal zero, resulting in a product that ultimately integrates to zero across symmetric bounds spanning negative to positive pi.Understanding how cosine behaves, particularly its periodic properties and zero crossings, helps to predict the integration result. It also hints why, in many symmetrical cases like this one, results turn out to be zero, highlighting symmetry in trigonometric functions.

XY-Plane

The \( xy \)-plane is a fundamental concept in integral calculus because it functions as the reference plane above which we calculate the volume of solids. In this exercise, the surface \( z = \cos x \cos y \) extends above and below the \( xy\)-plane.The plane is essentially defined where \( z = 0 \), and the intersection of the surface \( z = \cos x \cos y \) with the \( xy\)-plane determines certain symmetry properties and potential regions of interest. Here, integrating the function above and below the plane reveals a symmetry, essentially showing that the positive and negative volumes cancel out.Using the plane as a reference is critical to understanding whether an integral will yield a positive or zero volume. This also ties back to why, in this problem, the resulting volume between bounds is zero, due to the balanced positive and negative contributions of \( \cos x \cos y \) over the defined region.