Question

Two surfaces \(f_{1}(x, y)\) and \(f_{2}(x, y)\) and a region \(R\) in the \(x, y\) plane are given. Set up and evaluate the double integral that finds the volume between these surfaces over \(R\). \(f_{1}(x, y)=\sin x \cos y, f_{2}(x, y)=\cos x \sin y+2\); \(R\) is the triangle with corners \((0,0),(\pi, 0)\) and \((\pi, \pi)\).

Short answer

The volume between the surfaces is given by: \(2\pi^2 - \pi\).

Step-by-step solution

Define the Problem

We need to find the volume between two surfaces \(f_1(x, y) = \sin x \cos y\) and \(f_2(x, y) = \cos x \sin y + 2\) over a triangular region \(R\) with vertices at \((0,0)\), \((\pi, 0)\), and \((\pi, \pi)\). This involves setting up and evaluating a double integral of the difference between the surfaces over the region \(R\).

Set up the Integral

The volume between the surfaces is given by the double integral \( \int\int_R (f_2(x, y) - f_1(x, y)) \, dA \). Therefore, \( V = \int_{0}^{\pi} \int_{0}^{x} (\cos x \sin y + 2 - \sin x \cos y) \, dy \, dx \). The inner integral integrates with respect to \(y\) from 0 to \(x\), reflecting the boundaries of the triangular region.

Compute Inner Integral

Evaluate \( \int_{0}^{x} (\cos x \sin y + 2 - \sin x \cos y) \, dy \). Compute this as three separate integrals: 1. \( \int_{0}^{x} \cos x \sin y \, dy = \cos x [-\cos y]_0^x = \cos x (-\cos x + 1) \)2. \( \int_{0}^{x} 2 \, dy = [2y]_0^x = 2x \)3. \( \int_{0}^{x} -\sin x \cos y \, dy = -\sin x [\sin y]_0^x = -\sin x \sin x = -\sin^2 x \)Combine results: \((\cos x (-\cos x + 1) + 2x - \sin^2 x)\).

Integrate with Respect to x

Substitute the result from Step 3 into the outer integral:\[ \int_{0}^{\pi} (\cos x (1 - \cos x) + 2x - \sin^2 x) \, dx \]This simplifies to \( \int_{0}^{\pi} (\cos x - \cos^2 x + 2x - \sin^2 x) \, dx \). Notice that \( -\cos^2 x - \sin^2 x = -1 \), simplifying the integral to:\[ \int_{0}^{\pi} (\cos x - 1 + 2x) \, dx \].Compute each integral separately:1. \( \int_{0}^{\pi} \cos x \, dx = [\sin x]_0^\pi = 0 \)2. \( \int_{0}^{\pi} -1 \, dx = -[x]_0^\pi = -\pi \)3. \( \int_{0}^{\pi} 2x \, dx = [x^2]_0^\pi = 2\pi^2 \)Sum them: \(0 - \pi + 2\pi^2 = 2\pi^2 - \pi \).

Final Calculation and Solution

Combine the results of the outer integrals to find the total volume: \( V = 2\pi^2 - \pi \). This is the integral result representing the volume of the region between the given surfaces.

Key concepts

Volume Between Surfaces

To find the volume between two surfaces, you need to understand how to set up a double integral. The given problem involves surfaces described by functions: \( f_1(x, y) = \sin x \cos y \) and \( f_2(x, y) = \cos x \sin y + 2 \). To find the volume between these surfaces over a specified region, you calculate:

• The difference between the two functions, \( f_2(x, y) - f_1(x, y) \), which gives the height of the volume at each point \( (x, y) \).
• A double integral that sums up all these small volumes over the region \( R \).

The generic expression for this volume is:\[ V = \int\int_R (f_2(x, y) - f_1(x, y)) \, dA \] In practical terms, this means breaking down the task into manageable pieces, such as finding inner integrals and then outer integrals. By following the steps given in the solution, you can see how the problem is systematically broken down and solved.

Triangular Region

The region \( R \) over which you integrate is critical to understanding the setup of a double integral. Here, \( R \) is defined as a triangular region with vertices at \( (0,0) \), \( (\pi, 0) \), and \( (\pi, \pi) \). This triangular shape determines the limits of integration for \( x \) and \( y \):

• For the inner integral, \( y \) ranges from 0 to \( x \), reflecting that the length in the \( y \)-direction depends on your current position \( x \) along the triangle's base.
• For the outer integral, \( x \) simply ranges from 0 to \( \pi \), covering the full horizontal span of the triangle.

By accounting for the limits set by the triangle's shape, one can set up the integral reflecting the area of interest in this specific problem. The triangular region effectively dictates how we construct the bounds in the integral expression, ensuring the calculated volume only pertains to the region of interest.

Evaluate Integral

Evaluating an integral can often be broken down into simpler steps, especially when dealing with double integrals. After setting up the integral for the given problem, evaluating it consists of two main parts:

• First, compute the inner integral:\[ \int_{0}^{x} (\cos x \sin y + 2 - \sin x \cos y) \, dy \]This integral requires handling each term separately, leading to sub-integrals: 1. \( \int_{0}^{x} \cos x \sin y \, dy \)2. \( \int_{0}^{x} 2 \, dy \)3. \( \int_{0}^{x} -\sin x \cos y \, dy \)
• Second, substitute the result of the inner integral into the outer integral and evaluate:\[ \int_{0}^{\pi} (\cos x - 1 + 2x) \, dx \]

This structured approach simplifies complex expressions into manageable computations, eventually leading to the calculation of the exact volume between the surfaces.

Definite Integral

In this exercise, a definite integral is used to determine the volume enclosed between the two given surfaces within specific bounds. The limits of integration are defined by the triangular region and are set in terms of real numbers:

• The inner integral has its limits in terms of \( y \) going from 0 to \( x \).
• The outer integral limits \( x \) from 0 to \( \pi \).

A definite integral translates a mathematical function over an interval into an exact number, which is the value needed to understand the volume in this context. By substituting numerical bounds directly into the integral, you end up with a numeric result, such as the final answer \( V = 2\pi^2 - \pi \). This definite integral provides not only the limit values but also confidence in knowing the solution conforms to the given constraints of the problem.