Problem 12 - Thomas Calculus

Question

In Exercises \(1-14,\) evaluate the iterated integral. $$\int_{\pi}^{2 \pi} \int_{0}^{\pi}(\sin x+\cos y) d x d y$$

Step-by-Step Solution

Answer

The iterated integral evaluates to \( 2\pi \).

Step 1: Understand the Iterated Integral

The given iterated integral is \( \int_{\pi}^{2 \pi} \int_{0}^{\pi} (\sin x + \cos y) \, dx \, dy \). This means we first evaluate the integral with respect to \( x \), and then with respect to \( y \).

Step 2: Integrate with Respect to x

To evaluate the inner integral \( \int_{0}^{\pi} (\sin x + \cos y) \, dx \), we treat \( \cos y \) as a constant because we are integrating with respect to \( x \). The integral becomes:\[ \int_{0}^{\pi} \sin x \, dx + \int_{0}^{\pi} \cos y \, dx. \]Since \( \cos y \) is constant with respect to \( x \), its integral over \( x \) is simply \( \cos y \cdot x \) evaluated from 0 to \( \pi \). Now, we solve the two parts.

Step 3: Solve Each Part of the Integral

1. \( \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) + \cos(0) = 1 + 1 = 2. \)2. \( \int_{0}^{\pi} \cos y \, dx = \cos y \cdot [x]_{0}^{\pi} = \cos y \cdot \pi. \)

Step 4: Combine the Results of Inner Integral

Add the results from Step 3:\[ 2 + \pi \cos y. \]This expression is the result of the inner integral: \( \int_{0}^{\pi} (\sin x + \cos y) \, dx = 2 + \pi \cos y. \)

Step 5: Integrate with Respect to y

Now, evaluate the outer integral: \( \int_{\pi}^{2\pi} (2 + \pi \cos y) \, dy \). We split this into two separate integrals:1. \( \int_{\pi}^{2\pi} 2 \, dy = 2[y]_{\pi}^{2\pi} = 2(2\pi - \pi) = 2\pi. \)2. \( \int_{\pi}^{2\pi} \pi \cos y \, dy = \pi \sin y \bigg|_{\pi}^{2\pi} = \pi(\sin(2\pi) - \sin(\pi)) = 0. \)

Step 6: Combine the Results of Outer Integral

Add the results from Step 5:\[ 2\pi + 0 = 2\pi. \]

Step 7: Conclusion

The value of the iterated integral \( \int_{\pi}^{2 \pi} \int_{0}^{\pi} (\sin x + \cos y) \, dx \, dy \) is \( 2\pi. \)

Key Concepts

Integration Techniques

Integration techniques are various methods used to solve integrals. In this exercise, we focused on evaluating an iterated integral, which involves multiple integration layers. Initially, we integrated with respect to one variable, in this case, \(x\), while treating \(y\) as a constant, and then integrated with respect to another variable, \(y\).
To tackle such problems, it's important to identify the order of integration. This is often given in the problem as part of the notation. Remember that proper execution of each step is crucial for accuracy.

• Always take note which variable you are integrating first.
• Treat other variables as constants when not being integrated.
• Split and integrate each part of the expression separately if needed.

By understanding these strategies, problems involving iterated integrals become much more manageable.

Trigonometric Integrals

Trigonometric integrals involve integrating functions with trigonometric components, such as \(\sin(x)\) and \(\cos(y)\), like in our exercise. These functions often appear in calculus problems, and familiarity with their integration formulas is quite beneficial.
For instance, an integral of \(\sin x\) results in \(-\cos x\), while for \(\cos x\), it becomes \(\sin x\). In iterated integrals, it helps to:

• Recognize the relevant trigonometric identities.
• Apply constants based on integration intervals.
• Be mindful of periodic properties, which can simplify calculations.

Grasping these elements ensures you can handle trigonometric integrals efficiently.

Multiple Integrals

Multiple integrals are used to compute volumes and other multi-dimensional concepts. They extend the idea of iterated integrals by dealing with layers of integration for each dimension. In our exercise, we had a double integral, indicating two layers: first with \(x\), followed by \(y\).
These integrals can be represented in different orders, depending on the bounds and what is being calculated. Some helpful strategies include:

• Visualizing problems in terms of geometry for better intuition.
• Setting up bounds accurately, as they determine integration limits.
• Using symmetry of the function if applicable, to simplify the process.

Understanding multiple integrals enables tackling areas and volumes in higher dimensions beyond simple integration.

Calculus Problem Solving

Calculus problem solving involves applying mathematical concepts and techniques to analyze and solve problems. With iterated integrals, this includes a solid understanding of calculus fundamentals, careful execution of integration steps, and applying rules effectively.
When solving calculus problems:

• Break the problem into smaller, more manageable parts.
• Do a detailed step-by-step integration for clarity and accuracy.
• Verify solution consistency by checking against known results or different intervals.

Effective problem solving also means practicing a variety of problems to be adaptable across different contexts in calculus.