Problem 12 - University Calculus: Early Transcendentals

Question

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 value of the iterated integral is \(2\pi\).

Step 1: Identify the Order of Integration

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

Step 2: Integrate with Respect to x

The inner integral is \( \int_{0}^{\pi} (\sin x + \cos y) \, dx \). Since \( \cos y \) is constant with respect to \( x \), we can express this as \( \int_{0}^{\pi} \sin x \, dx + \int_{0}^{\pi} \cos y \, dx \). This separates into two integrals: \( \int_{0}^{\pi} \sin x \, dx \) and \( \cos y \int_{0}^{\pi} 1 \, dx \).

Step 3: Evaluate \( \int_{0}^{\pi} \sin x \, dx \)

The integral of \( \sin x \) is \( -\cos x \). Thus, \( \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \).

Step 4: Evaluate \( \int_{0}^{\pi} \cos y \, dx \)

Since \( \cos y \) is a constant with respect to \( x \), we integrate the constant \( \cos y \) over \( x \) from 0 to \( \pi \), which gives: \[ \cos y \int_{0}^{\pi} 1 \, dx = \cos y [x]_0^{\pi} = \cos y (\pi - 0) = \pi \cos y \].

Step 5: Combine the Results for the Inner Integral

Combining the results of Steps 3 and 4, we have: \[ \int_{0}^{\pi} (\sin x + \cos y) \, dx = 2 + \pi \cos y \].

Step 6: Integrate with Respect to y

Now integrate the result from Step 5 with respect to \( y \): \( \int_{\pi}^{2\pi} (2 + \pi \cos y) \, dy \). This separates into two integrals: \( \int_{\pi}^{2\pi} 2 \, dy \) and \( \pi \int_{\pi}^{2\pi} \cos y \, dy \).

Step 7: Evaluate \( \int_{\pi}^{2\pi} 2 \, dy \)

\( \int_{\pi}^{2\pi} 2 \, dy = 2[y]_{\pi}^{2\pi} = 2(2\pi - \pi) = 2\pi \).

Step 8: Evaluate \( \pi \int_{\pi}^{2\pi} \cos y \, dy \)

The integral of \( \cos y \) with respect to \( y \) is \( \sin y \). So, \( \pi \int_{\pi}^{2\pi} \cos y \, dy = \pi [\sin y]_{\pi}^{2\pi} = \pi(\sin(2\pi) - \sin(\pi)) = \pi(0-0) = 0 \).

Step 9: Combine the Results for the Outer Integral

The result of the outer integral is \( 2\pi + 0 = 2\pi \). Therefore, the value of the iterated integral is \( 2\pi \).

Key Concepts

Order of Integration

In calculus, when dealing with iterated integrals, the *order of integration* is crucial. Understanding the sequence of integration helps us approach the problem correctly. For example, in our exercise, we evaluated the iterated integral\[\int_{\pi}^{2 \pi} \int_{0}^{\pi} (\sin x + \cos y) \, dx \, dy\]The order in this integral indicates that we should first integrate with respect to \( x \), followed by \( y \).
This sequence is vital because it affects how you see and solve the problem. If the order is reversed, your approach and calculations will differ and might lead to a different result. Keeping track of the integration order helps in simplifying the integration process and maintaining the correct logic throughout the solution.

Integration with Respect to Variables

When evaluating iterated integrals, each variable integration requires special attention. The process often involves handling one variable, while considering others as constants. Let's consider the example integral with its separate terms,\[\int_{0}^{\pi} (\sin x + \cos y) \, dx\]Since \( \cos y \) acts as a constant in terms of \( x \), we split the integral into:

• \( \int_{0}^{\pi} \sin x \, dx \)
• \( \cos y \int_{0}^{\pi} 1 \, dx \)

This separation allows us to handle each part independently, simplifying calculations. With this approach, we easily compute specific parts and later, combine them to achieve the final result. Understanding how to treat variables as constants during integration is key to solving iterated integrals efficiently.

Evaluating Definite Integrals

Definite integrals involve evaluating an integral over specific limits. The calculated values represent the area under the curve for set intervals. For instance, in our integral solution:\[\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = 2\]here, we calculate the definite integral for \( \sin x \) over the range from 0 to \( \pi \).
The process involves finding an antiderivative and then applying the limits to find the exact value. We reached the result by substituting the upper and lower limits of \( x \), which shows the net accumulated area over that interval. Being precise in evaluations, especially while substituting the limits, ensures accuracy in the final result.

Trigonometric Functions Integration

Integrating trigonometric functions is a frequent task in calculus. Understanding their antiderivatives is essential. In our example, the integration involved both \( \sin x \) and \( \cos y \). Here, knowing that:

• The antiderivative of \( \sin x \) is \( -\cos x \)
• The antiderivative of \( \cos y \) is \( \sin y \)

These relationships helped simplify the integrals:

• \( \int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} \)
• \( \int_{\pi}^{2\pi} \cos y \, dy = [\sin y]_{\pi}^{2\pi} \)

In this scenario, applying these antiderivatives, combined with boundary conditions, led to simplified calculations. Mastery of trigonometric functions and their integrals is important in many calculus problems, directly affecting the outcome of complex integral evaluations.