Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the indicated double integral over \(R\). $$ \iint_{R}\left(x^{2}+y^{2}\right) d A ; R=\\{(x, y):-1 \leq x \leq 1,0 \leq y \leq 2\\} $$

Step-by-Step Solution

Answer

The value of the double integral is \( \frac{20}{3} \).

Step 1: Identify the Region of Integration

The region of integration, \( R \), is given as \( \{-1 \leq x \leq 1, 0 \leq y \leq 2\} \). This region is a rectangle in the xy-plane.

Step 2: Set up the Double Integral

The double integral over the rectangular region \( R \) can be expressed by iterating two single integrals. We'll choose to integrate with respect to \( y \) first, and then \( x \). Thus, the integral becomes \( \int_{-1}^{1} \int_{0}^{2} (x^2 + y^2) \, dy \, dx \).

Step 3: Integrate with respect to y

First, integrate \( x^2 + y^2 \) with respect to \( y \) from 0 to 2. The integral becomes:\[\int_{0}^{2} (x^2 + y^2) \, dy = \left[ x^2y + \frac{y^3}{3} \right]_{0}^{2}.\]Evaluating this gives:\[x^2(2) + \frac{2^3}{3} - (x^2(0) + \frac{0^3}{3}) = 2x^2 + \frac{8}{3}.\]

Step 4: Integrate with respect to x

Now, integrate the result from Step 3 with respect to \( x \) from -1 to 1:\[\int_{-1}^{1} \left( 2x^2 + \frac{8}{3} \right) \, dx.\]This simplifies to two separate integrals:\[2 \int_{-1}^{1} x^2 \, dx + \frac{8}{3} \int_{-1}^{1} 1 \, dx.\]

Step 5: Evaluate Each Integral

Evaluate \( \int_{-1}^{1} x^2 \, dx \):\[\left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}.\]Next, evaluate \( \int_{-1}^{1} 1 \, dx \):\[\left[ x \right]_{-1}^{1} = 1 - (-1) = 2.\]

Step 6: Combine the Results

Substitute the evaluated integrals back into the expression from Step 4:\[2 \times \frac{2}{3} + \frac{8}{3} \times 2 = \frac{4}{3} + \frac{16}{3} = \frac{20}{3}.\]

Step 7: Conclude the Solution

The value of the double integral \( \iint_{R} (x^2 + y^2) \, dA \) over the region \( R \) is \( \frac{20}{3} \).

Key Concepts

Region of Integration

In the context of double integrals, the **region of integration** refers to the specific area over which we integrate a given function. For the provided exercise, the region is defined as a rectangle in the xy-plane denoted by \(R = \{-1 \leq x \leq 1, 0 \leq y \leq 2\}\). This essentially tells us that the bounds for \(x\) are from -1 to 1, and for \(y\) are from 0 to 2.
Breaking it down further:

• The bounds for the **x-axis** are from -1 to 1, meaning our integration is happening in a horizontal direction between these two limits.
• The bounds for the **y-axis** are from 0 to 2, implying a vertical restriction between these values.

Visualizing this, picture a rectangle on a 2D plane where the bottom-left corner is at \((-1,0)\) and the top-right corner is at \((1,2)\). This rectangle becomes the map or frame within which we will perform our integration.

Iterated Integrals

**Iterated integrals** are a method we use to compute double integrals by breaking them down into successive single integrals. Essentially, it involves integrating one variable first while considering the others constant, and then following by integrating with respect to the next variable.
For our exercise, we first integrate with respect to \(y\) and then with respect to \(x\). Here’s how this looks:

• The inner integral \(\int_{0}^{2} (x^2 + y^2) \, dy\) involves integrating over \(y\) first. At this stage, \(x\) is treated as a constant.
• The outer integral \(\int_{-1}^{1} \, \) encompasses integrating over \(x\). Once we have the result from the first part, it completes the process by integrating this outcome with respect to \(x\).

Iterated integration helps manage complex multi-dimensional integrals by breaking them into more manageable one-dimensional problems. The order of integration can sometimes simplify the process significantly, depending on the function and the region.

Rectangle in xy-plane

The concept of a **rectangle in the xy-plane** is foundational in understanding regions of integration for double integrals, especially in elementary cases. Here, the rectangle \(R = \{-1 \leq x \leq 1, 0 \leq y \leq 2\}\) represents the specific domain in which we will evaluate our integral.
Key attributes of this rectangle:

• It extends horizontally from \(x = -1\) to \(x = 1\).
• It extends vertically from \(y = 0\) to \(y = 2\).

To visualize this concept better, imagine a box on a graph with these corner points serving as anchors. This physical area forms what is known as a **bounded region**, which provides the precise limits for our double integral. Understanding this geometric representation is critical as it directly influences how you set up the integral and calculate it.

Evaluating Integrals

**Evaluating integrals** refers to the process of calculating the actual numerical value of the integral. Once the region of integration and iterated integral setup are clear, it's time to perform the calculation.
Consider our exercise:

• Start by evaluating the inner integral with respect to \(y\): \(\int_{0}^{2} (x^2 + y^2) \, dy\) simplifies upon processing to \(2x^2 + \frac{8}{3}\).
• With the inner integral evaluated, focus on the outer integral with respect to \(x\): \(\int_{-1}^{1} (2x^2 + \frac{8}{3}) \, dx\).

By computing these steps sequentially, integrating each variable independently within its designated limits, we arrive at the final solution. For this problem, the outcome is \(\frac{20}{3}\). Carefully working through each step ensures accuracy and a clearer understanding of how these values are attained.