Calculate the double integral where is the region

Note that all three of these properties are really just extensions of properties of single integrals that have been extended to double integrals. The best way to do this is the graph the two curves. Here is a sketch. We got even less information about the region this time.

Now, there are two ways to describe this region. If we do this we can notice that the same function is always on the right and the same function is always on the left and so the region is,. That was a lot of work. Solution 2 This solution will be a lot less work since we are only going to do a single integral. So, the numbers were a little messier, but other than that there was much less work for the same result. As the last part of the previous example has shown us we can integrate these integrals in either order i.

In fact, there will be times when it will not even be possible to do the integral in one order while it will be possible to do the integral in the other order.

Also, do not forget about Calculus I substitutions. Students often just get in a hurry and multiply everything out after doing the integral evaluation and end up missing a really simple Calculus I substitution that avoids the hassle of multiplying everything out. We are going to hope that if we reverse the order of integration we will get an integral that we can do.

Even if we ignored that the answer would not be a constant as it should be. The best way to reverse the order of integration is to first sketch the region given by the original limits of integration. From the integral we see that the inequalities that define this region are,. Here is a sketch of that region. Here they are for this region. Now we could redo this example using a union of two Type II regions see the Checkpoint.

Consider the region bounded by the curves and in the interval Decompose the region into smaller regions of Type II. Redo Figure using a union of two Type II regions. As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration.

Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work. Reverse the order of integration in the iterated integral Then evaluate the new iterated integral. The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II.

Refer to Figure. We can see from the limits of integration that the region is bounded above by and below by where is in the interval By reversing the order, we have the region bounded on the left by and on the right by where is in the interval We solved in terms of to obtain. Consider the iterated integral where over a triangular region that has sides on and the line Sketch the region, and then evaluate the iterated integral by.

A sketch of the region appears in Figure. Evaluate the iterated integral over the region in the first quadrant between the functions and Evaluate the iterated integral by integrating first with respect to and then integrating first with resect to.

Sketch the region and follow Figure. We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those in Double Integrals over Rectangular Regions , but without the restriction to a rectangular region, we can now solve a wider variety of problems.

Find the volume of the solid bounded by the planes and. The solid is a tetrahedron with the base on the -plane and a height The base is the region bounded by the lines, and where Figure. First, consider as a Type I region, and hence. Now consider as a Type II region, so In this calculation, the volume is. Therefore, the volume is cubic units. Find the volume of the solid bounded above by over the region enclosed by the curves and where is in the interval.

Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general nonrectangular region, as stated in the next definition.

The area of a plane-bounded region is defined as the double integral. We have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by using double integrals, which can be very useful, as we will see in the later sections of this chapter. Find the area of the region bounded below by the curve and above by the line in the first quadrant Figure.

We just have to integrate the constant function over the region. Thus, the area of the bounded region is or. Find the area of a region bounded above by the curve and below by over the interval. We can also use a double integral to find the average value of a function over a general region. The definition is a direct extension of the earlier formula. If is integrable over a plane-bounded region with positive area then the average value of the function is.

Note that the area is. Find the average value of the function on the region bounded by the line and the curve Figure. First find the area where the region is given by the figure. We have. Find the average value of the function over the triangle with vertices. Express the line joining and as a function. An improper double integral is an integral where either is an unbounded region or is an unbounded function. For example, is an unbounded region, and the function over the ellipse is an unbounded function.

Hence, both of the following integrals are improper integrals:. In this section we would like to deal with improper integrals of functions over rectangles or simple regions such that has only finitely many discontinuities.

If is a bounded rectangle or simple region in the plane defined by and also by and is a nonnegative function on with finitely many discontinuities in the interior of then. It is very important to note that we required that the function be nonnegative on for the theorem to work. We consider only the case where the function has finitely many discontinuities inside. Consider the function over the region. First we plot the region Figure ; then we express it in another way. The other way to express the same region is.

As mentioned before, we also have an improper integral if the region of integration is unbounded. Suppose now that the function is continuous in an unbounded rectangle. If is an unbounded rectangle such as then when the limit exists, we have. The following example shows how this theorem can be used in certain cases of improper integrals. Evaluate the integral where is the first quadrant of the plane. The region is the first quadrant of the plane, which is unbounded. Thus, is convergent and the value is.

Evaluate the improper integral where. Notice that the integral is nonnegative and discontinuous on Express the region as and integrate using the method of substitution. In some situations in probability theory, we can gain insight into a problem when we are able to use double integrals over general regions.

Before we go over an example with a double integral, we need to set a few definitions and become familiar with some important properties. Consider a pair of continuous random variables and such as the birthdays of two people or the number of sunny and rainy days in a month. The joint density function of and satisfies the probability that lies in a certain region. Since the probabilities can never be negative and must lie between and the joint density function satisfies the following inequality and equation:.

The variables and are said to be independent random variables if their joint density function is the product of their individual density functions:. From the time they are seated until they have finished their meal requires an additional minutes, on average. What is the probability that a customer spends less than an hour and a half at the diner, assuming that waiting for a table and completing the meal are independent events?

Waiting times are mathematically modeled by exponential density functions, with being the average waiting time, as. Clearly, the events are independent and hence the joint density function is the product of the individual functions. We want to find the probability that the combined time is less than minutes. In terms of geometry, it means that the region is in the first quadrant bounded by the line Figure. Hence, the probability that is in the region is.

Since is the same as we have a region of Type I, so. Thus, there is an chance that a customer spends less than an hour and a half at the restaurant. Another important application in probability that can involve improper double integrals is the calculation of expected values.

First we define this concept and then show an example of a calculation. In probability theory, we denote the expected values and respectively, as the most likely outcomes of the events. The expected values and are given by. Using the first quadrant of the rectangular coordinate plane as the sample space, we have improper integrals for and The expected time for a table is. A similar calculation shows that This means that the expected values of the two random events are the average waiting time and the average dining time, respectively.

The joint density function for two random variables and is given by. Find the probability that is at most and is at least. Compute the probability. The region bounded by and as given in the following figure. Find the average value of the function on the region graphed in the previous exercise. Find the area of the region given in the previous exercise. The region bounded by as given in the following figure. Find the volume of the solid under the graph of the function and above the region in the figure in the previous exercise.

Find the volume of the solid under the graph of the function and above the region in the figure from the previous exercise. Let be the region bounded by the curves of equations and and the -axis. Explain why is neither of Type I nor II. The region is not of Type I: it does not lie between two vertical lines and the graphs of two continuous functions and The region is not of Type II: it does not lie between two horizontal lines and the graphs of two continuous functions and.

In the following exercises, evaluate the double integral over the region. Let be the region bounded by and the — and -axes. Let be the region bounded by and the -axis.

Answers may vary; b. Let be the region bounded by and. The region bounded by and is shown in the following figure.

Find the area of the region. Find the average value of the function on the triangular region with vertices and. In the following exercises, change the order of integration and evaluate the integral. The region is shown in the following figure. Evaluate the double integral by using the easier order of integration. The region is given in the following figure. Find the volume of the solid under the surface and above the region bounded by and.

Find the volume of the solid under the plane and above the region determined by and. Find the volume of the solid under the plane and above the region bounded by and.

Find the volume of the solid under the surface and above the plane region bounded by and. Let be a positive, increasing, and differentiable function on the interval Show that the volume of the solid under the surface and above the region bounded by and is given by. Let be a positive, increasing, and differentiable function on the interval and let be a positive real number.

Show that the volume of the solid under the surface and above the region bounded by and is given by. Find the volume of the solid situated in the first octant and determined by the planes.

Find the volume of the solid situated in the first octant and bounded by the planes. Find the volume of the solid bounded by the planes. Let and be the solids situated in the first octant under the planes and respectively, and let be the solid situated between.

Let be the solids situated in the first octant under the planes and respectively, and let be the solid situated between. Let be the solids situated in the first octant under the plane and under the sphere respectively. If the volume of the solid is determine the volume of the solid situated between and by subtracting the volumes of these solids.