Question

Evaluate the indicated double integral over \(R\). $$ \begin{array}{l} \iint_{R} \sin (x+y) d A \\ R=\\{(x, y): 0 \leq x \leq \pi / 2,0 \leq y \leq \pi / 2\\} \end{array} $$

Short answer

The value of the double integral is 2.

Step-by-step solution

Define the Double Integral

We are given the function \( f(x, y) = \sin(x+y) \). We need to evaluate the double integral \( \iint_{R} \sin (x+y) \, dA \) over the region \( R = \{(x, y): 0 \leq x \leq \pi/2, \, 0 \leq y \leq \pi/2 \} \).

Set Up the Iterated Integral

For the region \( R \), set up the iterated integral as \( \int_{0}^{\pi/2} \int_{0}^{\pi/2} \sin(x+y) \, dx \, dy \). The limits of integration for both \( x \) and \( y \) are from 0 to \( \pi/2 \).

Integrate with Respect to x

Integrate \( \int_{0}^{\pi/2} \sin(x+y) \, dx \). Change variables using \( u = x+y \), then \( du = dx \). The limits change as \( x = 0 \) to \( x = \pi/2 \), giving \( u = y \) to \( u = y + \pi/2 \). The integral becomes \( \int_{y}^{y+\pi/2} \sin(u) \, du\).

Evaluate the Integral with Respect to x

Compute \( \int \sin(u) \, du = -\cos(u) \). Evaluate from \( y \) to \( y + \pi/2 \):\[-\cos(y+\pi/2) + \cos(y) = \sin(y) + \cos(y)\].

Integrate with Respect to y

Now integrate with respect to \( y \): \( \int_{0}^{\pi/2} \big( \sin(y) + \cos(y) \big) \, dy \). This separates to two integrals: \( \int_{0}^{\pi/2} \sin(y) \, dy + \int_{0}^{\pi/2} \cos(y) \, dy \).

Evaluate the Integral of sin(y)

The integral \( \int \sin(y) \, dy = -\cos(y) \), and evaluated from 0 to \( \pi/2 \) gives \(-\cos(\pi/2) + \cos(0) = 0+1 = 1.\)

Evaluate the Integral of cos(y)

The integral \( \int \cos(y) \, dy = \sin(y) \), and evaluated from 0 to \( \pi/2 \) gives \( \sin(\pi/2) - \sin(0) = 1 - 0 = 1.\)

Sum the Results

Add the results from steps 6 and 7: \( 1 + 1 = 2 \). This gives the value of the double integral.

Key concepts

Iterated Integral

A double integral can often be broken down into two single integrals, one performed after the other. This is what we call an "iterated integral." In this exercise, the iterated integral went from a double integral over a region \( R \) to an integral over \( x \) followed by an integral over \( y \), or vice versa.

For example, the given double integral \( \iint_{R} \sin(x+y) \, dA \) is evaluated by first integrating with respect to \( x \), fixing the \( y \)-values, and then integrating with respect to \( y \). By choosing the order of integration, we can sometimes simplify the calculations significantly. Here, since the region \( R \) is a rectangle, either order will yield the same result.

• The setup of iterated integration involves choosing the integration limits for each variable.
• We use the inner integral to simplify the function with one variable and the outer integral to support the whole computation.

Change of Variables

Change of variables is a handy tool to simplify the integration process. Sometimes, an integral might appear daunting in its original form. By substituting variables with more convenient terms, the integral can be transformed into an easier one to evaluate.

In step 3 of the solution, the substitution \( u = x + y \) is used to replace \( x \). As a result, \(\sin(x+y)\) becomes \(\sin(u)\). This reduction can often make the function easier to integrate, particularly when the derivative of the substitution is straightforward like \( du = dx \).

• The substitution can sometimes result in updated integration limits, which occur naturally through the relation \( u = x+y \). Thus, \( x = 0 \) gives \( u = y \) and \( x = \pi/2 \) gives \( u = y + \pi/2 \).
• After substitution, the problem may become easier, allowing us to utilize simpler integration rules to solve the problem.

Integration Limits

Integration limits are crucial in defining the scope of the integration process. For a double integral, these limits outline the region over which integration is performed. It's important to correctly determine and understand these limits to correctly evaluate the integral.

In this exercise, the rectangular region \( R = \{(x, y): 0 \leq x \leq \pi/2, 0 \leq y \leq \pi/2 \}\) provides clearly defined limits. Hence, the iterated integral utilizes these bounds:

• First, integrate over \( x \) from 0 to \(\pi/2\).
• Then, integrate over \( y \) with the same limits.

This scenario illustrates how rectangular regions result in constant limits for both variables, simplifying the setup for integration. By listing these limits early, confusion is minimized and calculations proceed more smoothly.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric expressions, such as \( \sin(x) \), \( \cos(x) \), among others. These functions have well-known antiderivatives which can greatly simplify the integration process.

In our scenario, the function to integrate is \( \sin(x+y) \). Utilizing the antiderivative of \( \sin(x) \), which is \(-\cos(x)\), the problem becomes straightforward. Each segment of the integration applies these familiar rules.

• After the substitution, the integral \( \int \sin(u) \, du \) easily transforms to \(-\cos(u)\).
• Finally, evaluating this expression at our new bounds \( y \) and \( y + \pi/2 \) streamlines to \( \sin(y) + \cos(y)\).

These steps demonstrate how understanding the relationships between trigonometric functions and their antiderivatives enables efficient calculation of the integral. Breaking up integrals into known expressions can ease the work involved and lead to an exact solution.