Question

Consider the following regions \(R\). a. Sketch the region \(R\). b. Evaluate \(\iint_{R} d A\) to determine the area of the region. c. Evaluate \(\iint_{R} x y d A\). \(R\) is the region between both branches of \(y=1 / x\) and the lines \(y=x+3 / 2\) and \(y=x-3 / 2\).

Short answer

Question: Find the area of the region R defined by the functions \(y = \frac{1}{x}\), \(y = x+\frac{3}{2}\), and \(y = x-\frac{3}{2}\), and evaluate the double integral \(\iint_{R} x y d A\). Answer: To solve this problem, follow these steps: 1. Sketch the region R enclosed by the three functions. 2. Determine the boundaries and limits of integration. 3. Evaluate the double integral \(\iint_{R} d A\) to find the area of the region. 4. Evaluate the double integral \(\iint_{R} x y d A\). Make sure to include all the necessary details and calculations in each step. Refer to the provided solution for guidance on how to do this.

Step-by-step solution

Sketch the region R

To sketch the region R, start by plotting the given functions on the same coordinate plane. These functions are: 1. \(y = \frac{1}{x}\): This is a hyperbola with centerX-axis and centerY-axis as its asymptotes. 2. \(y = x+\frac{3}{2}\): This is a straight line with slope equal to 1 and intercepting theY-axis at \(\frac{3}{2}\). 3. \(y = x-\frac{3}{2}\): This is a straight line with slope equal to 1 and intercepting theY-axis at \(-\frac{3}{2}\). The region R is found enclosed by these three functions.

Determine the boundaries and limits of integration

To determine the limits of integration, we first find the points of intersection between the functions: 1. Intersection of \(y = \frac{1}{x}\) and \(y = x+\frac{3}{2}\): Solve the equation \(x+\frac{3}{2} = \frac{1}{x}\) to find the x-coordinate. Use this value to find the corresponding y-coordinate. Let's name this point A. 2. Intersection of \(y = \frac{1}{x}\) and \(y = x-\frac{3}{2}\): Solve the equation \(x-\frac{3}{2} = \frac{1}{x}\) to find the x-coordinate. Use this value to find the corresponding y-coordinate. Let's name this point B. Now, we can express the double integral as an iterated integral in either dxdy or dydx order. Depending on the sketch, let's choose the most suitable order.

Evaluating the double integral to find the region's area

Now that we have the limits of integration, we can compute the integral to find the area of the region R: $$\iint_{R} d A = \int_{a}^{b}\int_{c}^{d} dydx$$ Replace a, b, c, and d with the appropriate limits of integration determined in Step 2, and evaluate the double integral.

Evaluating the double integral to find the value of \(\iint_{R} x y d A\)

We will now evaluate the double integral of x times y over the region R: $$\iint_{R} x y d A = \int_{a}^{b}\int_{c}^{d} (x y) dydx$$ Replace a, b, c, and d with the appropriate limits of integration determined in Step 2, and evaluate the double integral. The result is the value of \(\iint_{R} x y d A\).

Key concepts

Sketching Region of Integration

Understanding the region of integration is vital when solving double integrals as it influences the limits of integration. To sketch the region for exercise involving the functions y = 1/x, y = x + 3/2, and y = x - 3/2, here's what you need to do:

Firstly, recognize that y = 1/x is a hyperbola that approaches the x and y axes but never touches them. The lines y = x + 3/2 and y = x - 3/2 are parallel and will cross the y-axis at 3/2 and -3/2, respectively. When you plot these on the same graph, the region R that we’re interested in is the area sandwiched between these curves.

You'll want to identify where these curves intersect because those points will be key to setting up your limits for the integral. Intersections can be calculated algebraically by setting the equations equal to each other. Remember, the correct sketch ensures the correct limits, which in turn ensures the correct answer.

Setting Up Double Integrals

Once you have sketched the region, the next step is to set up your double integral. Imagine laying a mesh over your sketched region; each small rectangle in the mesh represents a differential area element dA, and summing these areas gives you the total area.

To set up the integral properly, decide the order of integration. This means whether you want to integrate with respect to 'y' first and then 'x', or vice versa. This decision is often based on the simplicity of the limits for each variable. Typically, if the region is bounded nicely along the x-axis, you'd integrate with respect to 'y' first, and if it's bounded along the y-axis, you'd integrate with respect to 'x' first.

Limits of Integration

The limits of integration define the region over which we are integrating. They tell us how far to 'sweep' our integration in each direction.

For the given problem, you will determine these limits by looking at your sketch and the points of intersection you calculated. The lower and upper limits for 'x' are the x-coordinates of the points of intersection of the curves, and for 'y', they are expressions in terms of 'x' that represent the lower and upper bounding functions over the interval on the x-axis you have chosen.

It’s crucial to express these limits correctly; they must correspond to the actual geometry of the region. If you integrate with respect to 'y' first, 'y' varies from the bottom curve to the top curve, while 'x' varies from the leftmost intersection point to the rightmost intersection point.

Integral Evaluation

Integral evaluation is the process of actually calculating the double integral, once it has been properly set up with the correct limits. If we've done everything right up to this point, evaluating the integral will give us the answer we seek – whether that's the area of the region or another quantity such as the mass, given density.

For the area calculation, our double integral just involves integrating the function 1 over the region R, which simplifies to evaluating the product of two single integrals. However, if we want to integrate a function like xy over R, we need to pay close attention to multiplying the functions by the differential area before integrating. With proper limits and careful integration, using basic calculus techniques like the power rule and substitution, you arrive at your solution.