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Chapter 15: Problem 47

In Exercises \(47-56,\) sketch the region of integration, reverse the order of integration, and evaluate the integral. \begin{equation} \int_{0}^{\pi} \int_{x}^{\pi} \frac{\sin y}{y} d y d x \end{equation}

Short Answer

Expert verified
The integral evaluates to 2 after reversing the order of integration.

Step by step solution

01

Understand the Region of Integration

The given double integral is \( \int_{0}^{\pi} \int_{x}^{\pi} \frac{\sin y}{y} \, d y \, d x \). This means for each \( x \) from \( 0 \) to \( \pi \), \( y \) varies from \( x \) to \( \pi \). Consequently, the region R is bounded by \( y = x \), \( y = \pi \), \( x = 0 \), and \( x = \pi \). This region forms a triangle above the line \( y = x \) and below \( y = \pi \) in the first quadrant.
02

Sketch the Region of Integration

To sketch the region, draw two lines: \( y = x \) and \( y = \pi \), and vertical lines at \( x = 0 \) and \( x = \pi \). The region between these boundaries is the triangular region where integration occurs. This region's vertices are located at points \((0, 0), (\pi, \pi),\) and \((\pi, \pi)\).
03

Reverse the Order of Integration

To reverse the order, consider \( y \) to run from \( 0 \) to \( \pi \). Since \( y \) takes these values, \( x \) will start from \( 0 \) up to \( y \) (since for a fixed \( y \), \( x \) goes from the left side to the line \( y = x \)). Thus, the new limits are: \( \int_{0}^{\pi} \int_{0}^{y} \frac{\sin y}{y} \, dx \, dy \).
04

Simplify and Evaluate the Inner Integral

The inner integral, with respect to \( x \), becomes: \( \int_{0}^{y} 1 \, dx = x \Big|_{0}^{y} = y \). Therefore, the integral becomes \( \int_{0}^{\pi} y \frac{\sin y}{y} \, dy = \int_{0}^{\pi} \sin y \, dy \).
05

Evaluate the Remaining Single Integral

The integral \( \int_{0}^{\pi} \sin y \, dy \) is computed as \( -\cos y \Big|_{0}^{\pi} = [-\cos(\pi) + \cos(0)] = -(-1) + 1 = 2 \).
06

Conclude the Evaluation

The value of the integral after reversing the order of integration and evaluating is \( 2 \). Therefore, the initial double integral evaluates to 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reversing Order of Integration
Sometimes, reversing the order of integration in a double integral can simplify the calculation process. The idea is to change which variable is held constant and which variable is considered to range over a particular interval.
  • First, identify the current limits of integration. In our example, the limits are \( x \) from 0 to \( \pi \) and, for each \( x \), \( y \) from \( x \) to \( \pi \).
  • The new approach involves flipping these axes. So \( y \) will now go from 0 to \( \pi \), and for each \( y \), \( x \) will span from 0 to \( y \).
This method can be extremely useful when the inner integral involves a complex calculation that is simplified by reversing the integration order.
Region of Integration
Understanding the region of integration is crucial. It tells us the geometric shape over which integration occurs. In the problem at hand, we start with \( \int_{0}^{\pi} \int_{x}^{\pi} \frac{\sin y}{y} \, dy \, dx \).
This translates into a triangular region in the \( xy \)-plane:
  • The line \( y = x \) acts as a diagonal lower boundary.
  • The horizontal line \( y = \pi \) forms an upper boundary.
  • The vertical lines \( x = 0 \) and \( x = \pi \) enclose the region from left to right.
Visualizing this triangle ensures that when we reverse the order of integration, we accurately adjust our limits from subregions to reflect the entire primary region defined initially.
Trigonometric Functions
Trigonometric functions, such as \( \sin \) and \( \cos \), frequently appear in integrals due to their periodic properties and their relationship to geometry. In the given integral:
  • The function \( \frac{\sin y}{y} \) introduces a trigonometric term in the integrand, \( \sin y \), which can simplify upon evaluation as it becomes the integral of \( \sin y \).
  • The trigonometric functions are known for their unique properties like symmetry, periodicity, and well-established integrals, which makes them powerful tools in integration.
These properties are exploited in integration, notably when reversing the order of integration allows us to evaluate simpler trigonometric integrals.
Evaluation of Integrals
The primary goal when working with integrals, especially double integrals, is to evaluate them to arrive at a numerical result. For the integral \( \int_{0}^{\pi} \sin y \, dy \), this involves the fundamental theorem of calculus:
  • Compute the antiderivative of \( \sin y \), which is \( -\cos y \).
  • Apply the evaluated limits of integration, from 0 to \( \pi \), to the antiderivative.
This gives the result:\[ -\cos(\pi) + \cos(0) = 1 + 1 = 2 \]Once evaluated, this simpler single integral represents the original double integral, leading us to a final value of 2.

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Most popular questions from this chapter

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. \begin{equation} \int_{0}^{3 / 2} \int_{0}^{9-4 x^{2}} 16 x d y d x \end{equation}

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