Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 15-20, evaluate the given double integral by changing it to an iterated integral. \(\iint_{S}(x+y) d A ; S\) is the triangular region with vertices \((0,0),(0,4)\), and \((1,4)\).

Step-by-Step Solution

Answer

The value of the double integral is 2.

Step 1: Understand the Region of Integration

The given region of integration, \( S \), is a triangular region with vertices at \((0,0)\), \((0,4)\), and \((1,4)\). This triangle is oriented vertically from \(y=0\) to \(y=4\), and for any fixed \(y\), the \(x\)-value ranges from the line \(x=0\) to the line \(x=y/4\) from the slope of the line connecting \((0,0)\) and \((1,4)\).

Step 2: Set Up the Iterated Integral

Convert the double integral over region \( S \) into an iterated integral. Since \( y \) is described between two constants \(0\) and \(4\), and \( x \) varies between \(0\) and \(y/4\), we set up the integral as follows: \[ \int_{y=0}^{4} \int_{x=0}^{y/4} (x+y) \, dx \, dy. \]

Step 3: Integrate with respect to x

Integrate the function \( x+y \) with respect to \( x \) first, treating \( y \) as a constant. The integral of \( x \) is \( \frac{x^2}{2} \), and the integral of \( y \) is \( xy \). Thus, we have: \[ \int_{x=0}^{y/4} (x+y) \, dx = \left[ \frac{x^2}{2} + xy \right]_{0}^{y/4}. \]

Step 4: Evaluate the Inner Integral

Substitute \( x=y/4 \) and \( x=0 \) into \( \frac{x^2}{2} + xy \) and simplify: \[ \left( \frac{(y/4)^2}{2} + (y/4)y \right) - \left( \frac{0^2}{2} + 0 \right) = \frac{y^2}{32} + \frac{y^2}{4}. \] Combining terms results in \( \frac{y^2}{32} + \frac{8y^2}{32} = \frac{9y^2}{32} \).

Step 5: Integrate with respect to y

Now, integrate the final result \( \frac{9y^2}{32} \) with respect to \( y \) from \( y=0 \) to \( y=4 \): \[ \int_{y=0}^{4} \frac{9y^2}{32} \, dy = \frac{9}{32} \int_{0}^{4} y^2 \, dy = \frac{9}{32} \cdot \left[ \frac{y^3}{3} \right]_{0}^{4}. \]

Step 6: Evaluate the Outer Integral

Substitute \( y=4 \) and \( y=0 \) into \( \frac{y^3}{3} \) and calculate: \[ \frac{1}{3}(4^3) - \frac{1}{3}(0^3) = \frac{64}{3} - 0 = \frac{64}{3}. \] Thus, the result of the integral is \( \frac{9}{32} \times \frac{64}{3} = \frac{192}{96} = 2. \)

Key Concepts

Iterated Integrals

In multivariable calculus, a double integral involves integrating a function over a two-dimensional region. However, to evaluate this double integral practically, we often convert it into an iterated integral. An iterated integral is a sequence of single integrals that are evaluated one after the other. The process involves integrating with respect to one variable while treating the other variable as a constant, and then integrating the resulting expression with respect to the second variable.

This procedure significantly simplifies the calculations since single-variable integration is typically more manageable. When you convert a double integral over a region into an iterated integral, you choose an order of integration: either "dx dy" or "dy dx," depending on the region and the function's simplicity. The boundaries for these integrals are crucial, and they must accurately reflect the limits of your region of integration. Remember: the transition to iterated integrals allows you to tackle complex multivariable integration in a straightforward step-by-step manner.

Region of Integration

When working with double integrals, it's essential to have a clear understanding of the region of integration. This is the area over which the integration is performed. In our original exercise, the region is a triangular area with vertices at \(0,0\), \(0,4\), and \(1,4\).

To integrate effectively, particularly when switching to iterated integration, you have to ascertain the lines or curves that bound this region. In this case, for any fixed \(y\), \(x\) spans from \(x=0\) to \(x=y/4\), which is derived from the line between \(0,0\) and \(1,4\).

Being precise with the region's boundaries ensures that the integration is calculated over the correct area. It also affects the order of integration and how you set limits in the iterated integral. Visualizing this space or sketching it can be very helpful in aligning your integrals correctly.

Integration Techniques

Various techniques can be employed to solve integrals efficiently. For iterated integrals, you integrate either variable independently first, making sure to correctly handle all constants while integrating. In this exercise, integrating with respect to \(x\) first means treating \(y\) as a constant. So, when integrating \(x+y\), you handle it as two distinct terms: \(x\) and \(y\).

- Integrate the \(x\) term: The integral is \(\frac{x^2}{2}\).
- Integrate the constant \(y\) term: The integral of a constant \(y\) w.r.t \(x\) is \(xy\).

After calculating these separate integrals, substitute the boundary values and simplify. Next, proceed with integrating this result concerning \(y\). Each step should adhere to proper integration rules and algebraic simplification. Remember, a critical part of successful integration is following the correct steps and interpreting the result correctly after each integration.

Mathematical Problem Solving

Mathematical problem solving with double integrals requires logical thinking and precision. It's essential to break down the problem into smaller steps, as shown in the exercise solution. Start by clearly understanding your region and reformulating the single intgerals.

Organize your approach:

• Define your region's boundaries accurately.
• Set up your iterated integral correctly based on these boundaries.
• Select the easiest order (dx dy or dy dx) to simplify calculations.
• Carry out integration step by step, using the correct techniques.

Verify your results at each stage. Proper arithmetic, correct substitution of limits, and thorough simplifications are crucial in obtaining the right answer. Double-check any variable substitutions and arithmetic simplifications to ensure accuracy. Consistent practice with different problems sharpens these problem-solving skills, enabling quicker and more accurate solutions for complex integrals.