Problem 17 - Calculus Early Transcendentals

Question

Evaluate each double integral over the region \(R\) by converting it to an iterated integral. $$\iint_{R}(x+2 y) d A ; R=\\{(x, y): 0 \leq x \leq 3,1 \leq y \leq 4\\}$$

Step-by-Step Solution

Answer

Question: Evaluate the double integral of the function f(x,y) = x + 2y over the region R, given the limits 0 ≤ x ≤ 3 and 1 ≤ y ≤ 4. Answer: 81

Step 1: Identify the limits of integration

The given region, \(R\), is defined by the limits of \(x\) and \(y\): $$ 0 \leq x \leq 3 \\ 1 \leq y \leq 4 $$

Step 2: Setup the inner integral

As a first step, let's set up the inner integral, integrating with respect to \(x\): $$\int_{0}^{3}(x+2y)dx$$

Step 3: Evaluate the inner integral

Now let's evaluate the integral: $$ \int_{0}^{3}(x+2y)dx = \left[\frac{1}{2}x^2 + 2yx\right]_{0}^{3} = \left(\frac{1}{2}(3)^2 + 2(3)y\right) - \left(\frac{1}{2}(0)^2 + 2(0)y\right) = \left(\frac{9}{2} + 6y\right) $$

Step 4: Setup the outer integral

Now, let's set up the outer integral, integrating with respect to \(y\): $$\int_{1}^{4}\left(\frac{9}{2}+6y\right)dy$$

Step 5: Evaluate the outer integral

Finally, let's evaluate this integral: $$ \int_{1}^{4}\left(\frac{9}{2}+6y\right)dy = \left[\frac{9}{2}y + 3y^2\right]_{1}^{4} = \left(\frac{9}{2}(4) + 3(4)^2\right) - \left(\frac{9}{2}(1) + 3(1)^2\right) = 81 $$ The value of the double integral is \(81\).

Key Concepts

Iterated Integral

Double integrals involve integrating over a two-dimensional region, and one common method is to transform the double integral into an iterated integral. An iterated integral means performing integration in a sequence—one variable at a time. Let's break down how this works in simpler terms, using the problem as an example.

First, for the expression \(x+2y\), the integrals can be split to handle \(x\) and \(y\) separately within specified bounds. We integrate the inner integral, which keeps one variable fixed and assesses the integral with respect to the other variable.

• The inner integral is with respect to \(x\), and its limits are from 0 to 3. This essentially 'slices' the overall problem one direction at a time.
• Then, the outer integral takes this solution and integrates with respect to \(y\) over its limits, which are from 1 to 4. This tackles the problem in the other direction, culminating in the complete solution.

By performing iterated integration, we systematically break down the complex task of a double integral into manageable steps, providing a clear path to solving these problems.

Region of Integration

Understanding the region of integration is crucial in solving double integrals, as it defines the 'territory' over which you are integrating. This region is typically denoted by \(R\), and in our problem, it comprises all the pairs \((x, y)\) satisfying given boundary conditions.

When we express region \(R\) as \{(x, y): 0 \leq x \leq 3, 1 \leq y \leq 4\}\, this tells us that:

• The values of \(x\) range from 0 to 3, covering the horizontal span of the region.
• The values of \(y\) range from 1 to 4, which illustrates the vertical extent.

Visually, in the xy-plane, this forms a rectangular area bound by these limits. Grasping the region's boundaries allows you to understand exactly where your double integral will apply, and sets the necessary framework for evaluating the limits of integration precisely.

Limits of Integration

The limits of integration in a double integral mark the boundaries for each variable within the iterated integral. They delineate the start and end points for both the inner and outer integrals. In our integral:

• For the inner integral with respect to \(x\), the integration limits are from 0 to 3. These bounds originate from the definition of region \(R\) and guide how we integrate along the horizontal dimension.
• For the outer integral, integrating with respect to \(y\), the limits range from 1 to 4, reflecting the vertical dimension of our region of interest.

These limits are crucial as they ensure the integration happens over the correct section of the graph. Accurately applying these limits is essential for computing the correct area or volume represented by the double integral.