Problem 28 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch is helpful. \(\iint_{R}(x+y) d A ; R\) is the region in the first quadrant bounded by \(x=0, y=x^{2},\) and \(y=8-x^{2}\).

Step-by-Step Solution

Answer

Question: Evaluate the double integral of the function f(x, y) = x + y over the region R bounded by x = 0, y = x^2, and y = 8 - x^2. Answer: The value of the double integral is 33⅓.

Step 1: Determine the limits of integration.

To find the limits of integration, we first need to sketch the region R described by the given inequalities: \(x=0, y=x^2\) and \(y=8-x^2\). This region is in the first quadrant and enclosed by these three curves. Intersection points for \(y=x^2\) and \(y=8-x^2\): \(x^{2}=8-x^{2} \implies 2x^{2}=8 \implies x^2=4 \implies x=\pm2\). Since we are in the first quadrant, we have x = 2. Now let's find the corresponding y-coordinate: \(y=x^{2} \implies y=2^{2} =4\). So, the intersection point is \((2, 4)\). Now let's set the limits of integration: x limits: Since the region is in the first quadrant and bounded from the left by the y-axis, x ranges from 0 to 2. y limits: First, let's look at the lower curve, y=x^2. Using the inverse relation, x = sqrt(y). Second, the upper curve is given by y = 8 - x^2 and the inverse relation is x = sqrt(8-y). Therefore, y ranges from x^2 to 8-x^2. Now we can set up the double integral.

Step 2: Evaluate the double integral.

Now, we will set up the double integral and evaluate it. \(\iint_{R}(x+y) dA = \int_0^2 \int_{x^2}^{8-x^2} (x+y) dy dx\). First, we will integrate with respect to y: \(\int_0^2 [\int_{x^2}^{8-x^2} (x+y) dy] dx = \int_0^2 [xy + \frac{1}{2}y^2]_{y=x^2}^{y=8-x^2} dx\). After substituting the y limits: \(\int_0^2 [(x(8-x^2) + \frac{1}{2}(8-x^2)^2) - (x^3 + \frac{1}{2}(x^2)^2)] dx\). Now, we integrate with respect to x: \(= [\frac{8}{2}x^2 - \frac{x^4}{4} + \frac{1}{6}(8-x^2)^3 - \frac{x^4}{4}+\frac{1}{6}(x^2)^3]_0^2\) Now we substitute in the x limits: \(= [\frac{8}{2}(2)^2 - \frac{(2)^4}{4} + \frac{1}{6}(8-(2)^2)^3 - \frac{(2)^4}{4}+\frac{1}{6}((2)^2)^3]\) \(= [16 - 4 + \frac{1}{6}(4)^3 - 4 +\frac{1}{6}(4)^3]\) The final answer is: \([16 - 4 + \frac{64}{6} - 4 +\frac{64}{6}] = 12+\frac{128}{6} = 12+ 21\frac{1}{3} = 33\frac{1}{3}\).

Key Concepts

Limits of Integration

Understanding the limits of integration is crucial when dealing with double integrals. These limits define the region over which we are integrating, and they can vary in complexity depending on the shape of the region. In our exercise, we are given a region in the first quadrant, and it's our job to determine between what boundaries the variables x and y will vary. To visualize this, a sketch can be incredibly helpful: draw the curves described by the equations and the axes to see the region R where the functions intersect.

Finding the Intersection Points

The initial step involves establishing where the curves intersect each other, as these points will help us determine the limits for x and y. Remembering to keep within the first quadrant, we find the intersection at the point (2,4).

Setting the x and y Limits

Once the intersection is known, the x limits are found by observing the horizontal span of the area, from the y-axis to the most rightward point of the region. For our problem, this is from x = 0 to x = 2. For y, it’s a bit trickier; y varies between the lower and upper functions, from y = x^2 to y = 8 - x^2 for any given x within our x limits. It’s also important to note that the y limits are functions of x, which will affect how we evaluate the integral.

First Quadrant Region

Working with regions located in the first quadrant has some convenient implications. Firstly, because the first quadrant is where both x and y values are positive, it simplifies some of our work. The area we are dealing with is bounded by non-negative axes and non-negative functions.

Advantages in the First Quadrant

Being in the first quadrant means there is no need to worry about absolute values or negative integrals, which generally makes the arithmetic simpler. Moreover, any real-valued function that you are integrating is typically easier to visualize in this quadrant, aiding in understanding the geometry of the problem.

Implications for Our Problem

For the given exercise, knowing that our entire integration region lies in the first quadrant assures us that we will only encounter real, positive intersection points and that the pictorial representation of the region will be easier to interpret. This justifies our choice of intersection point (x=2), aligning with both the given boundaries and the nature of the first quadrant.

Integration Evaluation

When we evaluate double integrals, we deal with the integration of one variable at a time. With the correct limits of integration in place, the actual computation involves two separate integration steps, each taking into account a different variable.

Step-by-Step Integration

Firstly, in our example, we tackle the integration with respect to y, treating x as a constant. After performing this step, we substitute the upper and lower boundaries for y as determined previously. The resulting expression simplifies to involve only x, which we then integrate with respect to the x limits.

Calculating the Final Values

The last step is to replace the x values with the boundaries we found during the first step. The final calculation should yield a numerical answer, which in our case was 33 and 1/3. Keep in mind that meticulous application of limits and careful simplification at each step are key to achieving the correct result.

Remember that an integral evaluates to the accumulated sum over a range—it's essential to understand that in the context of double integrals, we are summing over a two-dimensional region, hence the importance of correctly establishing and integrating over the x and y limits. This stepwise approach helps avoid common mistakes and ensures a thorough understanding of each phase of the integration process.