# Exercise And Solution On Application Of Double Integral Pdf

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- 14.2bE: Double Integrals Part 2 (Exercises)
- 13.E: Applications of Multiple Integration (Exercises)
- multiple integrals pdf

*Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section. Here is a list of all the sections for which practice problems have been written as well as a brief description of the material covered in the notes for that particular section.*

We give here the formula for calculation of the average value of a distributed function. Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website. These cookies do not store any personal information. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies.

## 14.2bE: Double Integrals Part 2 (Exercises)

Make a test function that calls the trapezoidal function in trapezoidal. Make a test function that calls the midpoint function in midpoint. Compute the error too. Another alternative would of course be to ask the user for the value of n. Compute the integral by hand and illustrate how the two numerical methods approximates the integral. Compare with the exact value. Analytically, the integral computes to 2. By hand, with the trapezoidal method, we get 1. Graphically Figure 10 , it is clear that the numerical approach will have to under-estimate the true result.

The area under the blue graph in Figure 10 corresponds to the "true" area under the graph of the integrand. The area under the red graph corresponds to what you get with the trapezoidal method and two intervals. By hand, with the midpoint method, we get 2. Graphically Figure 11 , we might see that the numerical approach will have to over-estimate the true result.

Again, the area under the blue graph in Figure 11 corresponds to the "true" area under the graph of the integrand. The area under the red graph corresponds to what you get with the midpoint method and two intervals. Analytically, the integral computes to zero. In this case, both numerical methods will correctly calculate the integral to zero even with just two intervals!

Graphically, we see that they arrive at zero for "different reasons". The trapezoidal method constructs both its trapezoids by use of the function integrand value at the midpoint of the whole interval. However, there the integrand crosses the x-axis, i. The "area" computed by the trapezoidal method becomes the zero area located "between" the green graph and the x-axis in Figure The midpoint method computes the areas of two rectangles red graph in Figure 12 , but since the signs of these equal areas differ, they add to zero.

Note that also the midpoint method will do an exact to within machine precision integration when the integrand is a straight line. This is so, since the errors from each rectangle will cancel. What happens? How must you change the test to make it useful?

How does the convergence rate test behave? Any need for adjustment? We may understand this by considering the new function, i.

This makes the numerical calculation more inaccurate than previously. To fix this problem, we have several options. One possibility is to relax the tolerance, but this is not very satisfactory.

After all, the calculation is supposed to be "exact" for a straight line. Adjustments to the code are made according to the remark below and hand calculations are written into the code. Remarks The convergence rate test fails. Exercise Rectangle methods The midpoint method divides the interval of integration into equal-sized subintervals and approximates the integral in each subinterval by a rectangle whose height equals the function value at the midpoint of the subinterval.

Instead, one might use either the left or right end of the subinterval as illustrated in Figure This defines a rectangle method of integration. The height of the rectangle can be based on the left or right end or the midpoint. Figure Illustration of the rectangle method with evaluating the rectangle height by either the left or right point. Make sure you test for height equal to left , right , and mid. See code above. This may of course be avoided, if desirable, by asking the user for epsilon.

From Figure 14 , we see that the stricter the value of epsilon, the larger the difference between the midpoint and the trapezoidal methods. For a given value for epsilon, the midpoint method comes within the tolerance with fewer intervals lower value of n than does the trapezoidal method. This should be expected, since the midpoint method is somewhat more accurate than the trapezoidal method. The true error can very seldom be computed since we do not know the exact answer to the computational problem , so one has to find other indicators of the error, such as the one here where the changes in the integral value, as the number of intervals is doubled, is taken to reflect the error.

Note that Wolfram alpha does give you an answer, but that answer is an approximation, it is not exact. This is because Wolfram alpha too uses numerical methods to arrive at the answer, just as you will in this exercise. Therefore, we are forced to compute the integral by numerical methods. Compute a result that is right to four digits.

Use ideas from Exercise Adaptive integration. Running the program, gives a printout to the screen showing how diff gets smaller and smaller each time. The final two lines of the printout read: The integral computes to: 2. Remember that since we have not computed the exact error, we can not guarantee that the "result is right to four digits". However, we have reason to believe that we "are close".

Typically, when it is known that the error measure used is not precise, the tolerance is made stricter. Exercise Revisit fit of sines to a function This is a continuation of Exercise Fit sines to straight line. See code given above. Save the plot to a file with name filename. Observe how the approximation improves. There are ways to fix this issue. Running the program also produces the plot seen in Figure Clearly, the approximation is no good at the endpoints of the interval.

Implement and test this rule. And what is programming? The integral computes to: 2.

## 13.E: Applications of Multiple Integration (Exercises)

OK, so let's do it. Related documents. Multiple Integrals Sol. Multiple integrals are a generalization of the definite integral to functions of more than one variable. We can now define the double integral of a real-valued function f x,y over more general regions in R2.

The graphical interpretation of the double integral will be that it is the volume of the part of space under the roof. Most of the previous results hold in this situation as well, but some techniques need to be extended to cover this more general case. The most common multiple integrals are double and triple integrals, involving two or three variables, respectively. Calculate double integrals over a rectangle R. No enrollment or registration.

## multiple integrals pdf

In this section we investigate double integrals and show how we can use them to find the volume of a solid over a rectangular region in the x y x y -plane. Many of the properties of double integrals are similar to those we have already discussed for single integrals. We begin by considering the space above a rectangular region R. The base of the solid is the rectangle R R in the x y x y -plane. We want to find the volume V V of the solid S.

What does this mean? In Exercises , evaluate the integral and subsequent iterated integral. Evaluate one of the iterated integrals to find the area.

You might temporarily think of this surface as representing physical topography—a hilly landscape, perhaps. What is the average height of the surface or average altitude of the landscape over some region? As with most such problems, we start by thinking about how we might approximate the answer. For reasonably nice functions this does indeed happen.

#### Chapter 4 : Multiple Integrals

Recall that area between two curves is defined as the integral of the top curve minus the bottom curve. This idea can be brought to three dimensions. We defined the volume between two surfaces as the double integral of the top surface minus the bottom surface. This can be written formally with the theorem below. Then the volume of the surface is equal to the double integrals:. Notice that all the typical properties of the double integral hold.

If we have a mass density function for a lamina thin plate , how does a double integral determine the mass of the lamina? Given a mass density function on a lamina, how can we find the lamina's center of mass? What is a joint probability density function? How do we determine the probability of an event if we know a probability density function? The following preview activity explores how a double integral can be used to determine the density of a thin plate with a mass density distribution. Recall that in single-variable calculus, we considered a similar problem and computed the mass of a one-dimensional rod with a mass-density distribution.

One difficult part of computing double integrals is determining the limits of integration, i. Changing the order of integration is slightly tricky because its hard to write down a specific algorithm for the procedure. We demonstrate this process with examples. The simplest region other than a rectangle for reversing the integration order is a triangle. You can see how to change the order of integration for a triangle by comparing example 2 with example 2' on the page of double integral examples. In this page, we give some further examples changing the integration order. We have also labeled all the corners of the region.

This may be as a consequence either of the shape of the region, or of the complexity of the integrand. Calculating the double integral in the new coordinate system can be much simpler. We use change of variables to simplify the integral. Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website.

Make a test function that calls the trapezoidal function in trapezoidal. Make a test function that calls the midpoint function in midpoint. Compute the error too.

*Hence, the total Riemann sum approximates the volume under the surface by the volume of a bunch of these thin boxes. Double integral Riemann sum.*

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## 5 Comments

Hans R.Exercise 1. Express the following double integrals over a rectangle R as products of single integrals f(x, y) dxdy as an iterated double integral if D is a region bounded What is the Remark: Divide region in example (c) into two parts.

MaccompportsundTo illustrate computing double integrals as iterated integrals , we start with the simplest example of a double integral over a rectangle and then move on to an integral over a triangle.

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