Problem 4 - Thomas Calculus

Question

In Exercises \(1-10,\) sketch the region of integration and evaluate the integral. $$ \int_{\pi}^{2 \pi} \int_{0}^{\pi}(\sin x+\cos y) d x d y $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Region of Integration

The integral \( \int_{\pi}^{2 \pi} \int_{0}^{\pi} (\sin x + \cos y) \, dx \, dy \) specifies a region of integration where \( x \) ranges from 0 to \( \pi \) and \( y \) ranges from \( \pi \) to \( 2\pi \). This forms a rectangular region in the \(xy\)-plane.

Step 2: Rearrange the Double Integral

The problem requires evaluating the double integral: \( \int_{\pi}^{2 \pi} \int_{0}^{\pi} (\sin x + \cos y) \, dx \, dy \). We first consider treating each function inside the integral separately to make calculations easier.

Step 3: Compute the Inner Integral with Respect to x

Compute \( \int_{0}^{\pi} (\sin x + \cos y) \, dx \). Since \( \cos y \) is constant with respect to \( x \), this becomes two separate integrals: \( \int_{0}^{\pi} \sin x \, dx + \int_{0}^{\pi} \cos y \, dx \).

Step 4: Evaluate \( \int_{0}^{\pi} \sin x \, dx \)

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

Step 5: Evaluate \( \int_{0}^{\pi} \cos y \, dx \)

Since \( \cos y \) is constant with respect to \( x \), it simply integrates to \( \cos y \cdot \pi \) over the interval: \( \pi \cos y \).

Step 6: Combine Results of Inner Integral

Combine results to get: \( \int_{0}^{\pi} (\sin x + \cos y) \, dx = 2 + \pi \cos y \).

Step 7: Compute the Outer Integral with Respect to y

Now evaluate \( \int_{\pi}^{2 \pi} (2 + \pi \cos y) \, dy \). This can be split into \( \int_{\pi}^{2 \pi} 2 \, dy + \int_{\pi}^{2 \pi} \pi \cos y \, dy \).

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

The integral yields \( [2y]_{\pi}^{2\pi} = 2(2\pi) - 2(\pi) = 4\pi - 2\pi = 2\pi \).

Step 9: Evaluate \( \int_{\pi}^{2 \pi} \pi \cos y \, dy \)

The integral \( \int_{\pi}^{2 \pi} \pi \cos y \, dy = \pi [\sin y]_{\pi}^{2\pi} = \pi (\sin(2\pi) - \sin(\pi)) = 0 \) as both \( \sin(\pi) \) and \( \sin(2\pi) \) equal 0.

Step 10: Combine Results of Outer Integral

Sum results from \( \int_{\pi}^{2 \pi} 2 \, dy \) and \( \int_{\pi}^{2 \pi} \pi \cos y \, dy \): \( 2\pi + 0 = 2\pi \).

Key Concepts

Region of Integration

In the context of double integrals, the region of integration refers to the area over which we perform the integration. For the given integral\[\int_{\pi}^{2 \pi} \int_{0}^{\pi} (\sin x + \cos y) \ dx \, dy\]the region of integration is determined by the limits of integration for the variables \(x\) and \(y\). Specifically:

• \(x\) ranges from 0 to \(\pi\)
• \(y\) ranges from \(\pi\) to \(2\pi\)
This forms a rectangular region in the \(xy\)-plane with vertices at \((0, \pi)\), \((\pi, \pi)\), \((0, 2\pi)\), and \((\pi, 2\pi)\). Understanding this region is critical as it gives context to the entire problem and shows where the function \(\sin x + \cos y\) is being evaluated.

Inner Integral

The inner integral in a double integral setup is the integral that is evaluated first. In this calculation, we first integrate with respect to \(x\):\[\int_{0}^{\pi} (\sin x + \cos y) \, dx\]This involves treating \(y\) as a constant. Let's break it down:

• Separate the terms: \(\int_{0}^{\pi} \sin x \, dx + \int_{0}^{\pi} \cos y \, dx\)
• The integral of \(\sin x\) over \(0\) to \(\pi\): given by \([-\cos x]_{0}^{\pi} = 2\)
• The integral of \(\cos y\) is simply \(\cos y \cdot \pi\) since \(\cos y\) is a constant with respect to \(x\)

Finally, the result of the inner integration is \(2 + \pi \cos y\). This result will then be used in the outer integral.

Outer Integral

After computing the inner integral, the result is integrated with respect to \(y\) in the outer integral:\[\int_{\pi}^{2 \pi} (2 + \pi \cos y) \, dy\]This becomes:

• \(\int_{\pi}^{2 \pi} 2 \, dy\) which evaluates to \(2(2 \pi - \pi) = 2\pi\)
• \(\int_{\pi}^{2 \pi} \pi \cos y \, dy\), which evaluates to \(\pi [\sin y]_{\pi}^{2\pi} = 0\) as \(\sin(\pi) = \sin(2\pi) = 0\)

The final result of the outer integral is \(2\pi + 0 = 2\pi\). This completes the evaluation of the original double integral.

Trigonometric Integration

Trigonometric integration is a key part of solving this double integral. It involves integrating functions such as \(\sin x\) and \(\cos y\), each having specific rules:

• The integral of \(\sin x\) from \(0\) to \(\pi\) is \([-\cos x]_{0}^{\pi} = -(-1) + 1 = 2\)
• The integral of \(\cos y\) with respect to \(y\) translates into considering the antiderivative \(\sin y\). Here, \(\int_{a}^{b} \cos y \, dy = \sin y\Big|_a^b\)

Understanding these integration rules helps in simplifying the process. These foundational techniques apply widely across calculus problems involving trigonometric functions and are crucial for solving such integrals efficiently.