Evaluating a double integral by converting from rectangular coordinates

 Goal: Evaluate a double integral by converting from rectangular coordinates to polar coordinates 

In the last post, we showed how to convert a double integral from rectangular coordinates to polar coordinates. The first step consists in sketching the region.

Example

Solution

The region is the set of circles comprised between the circle of radius 1 and the circle of radius 2. Let's start by sketching the region. The inferior limit of the set of circles is represented in red. It has 1 as radius. The superior limit is represented in blue. Its radius is 2. Note that x ≤ 0, therefore the region is located in the negative part of the plane of coordinates.

In polar coordinates, the radius of the region represented by the set of circles comprised between the red one and the blue one varies from 1 to 2. The θ angle varies from ℼ/2 to 3ℼ/2. Therefore, R is an annular region that can be represented by:

Practice

Double Integrals over a polar rectangular region

 Goal: Evaluate a double integral over a polar rectangular region

Definition

The double integral of a function f(r, θ) over a polar rectangular plane in the θ plane is defined as 

As in double integrals over a rectangular region, the double integral over a polar rectangular region can be expressed as in iterated integral in polar coordinates.

In polar coordinates dA is replaced by rdrdθ. The double integral of f(x,y) in rectangular coordinates can be expressed in polar coordinates by substitution. This is done by substituting x by rcosθ, y by rsinθ and dA by rdrdθ.

The properties for double integrals over rectangular coordinates apply also to the double integrals over polar coordinates.

Example 1

Solution

As we can see in the figure below, r =1 and r = 3 represent circles of radius r =1 and r =3 and 0≤ θ ≤ℼ covers the entire top half of the plane.

Example 2

Solution

The figure is similar to that in example 1 but with outer radius 2. Do it by yourself.

Practice

Double Integrals and Expected Values in Probability Theory

 Expected Values Values

In probability theory, we define the expected values E(x) and E(y) as the most likely outcomes of the events. They are given by the following formulas:

where S is the sample space of the random variables X and Y.

Example..

We resume the example from the last post:

What is the expected time for the events "waiting for a table" and "completing the meal"

Solution

Using the first quadrant of coordinate plane as as the sample space, let's call E(X) the expected time for "waiting for a table"  and E(Y) the expected time for "waiting for a table". Let's calculate E(X)and E(Y):

 

Applications of double integrals in Probability

One application of the double integrals over general regions is found in calculating the probability of two random variables. Let's start by defining some notions.

Density function of one variable

  Let's consider the following figure:

Let's consider the point (a, f(a)) of the curve. f(a) represents the probability density of this point. Likewise  f(b) represents the probability density corresponding to the point (b, f(b)). Any point of the curve  has a density probability. The area under curve between X = a and X = b represents the sum of all the probability densities of all the points of the curve between f(a) and f(b).. It represents the integral of the function f over the interval [a, b]. It represents also the probability that any variable between a and b falls between the range [a. b]. The function f is the probability density function of f over [a, b].

Joint Density function of two variables

The notion of probability density function of one variable can be extended to two variables. Since we deal with two variables, we use double integrals. The probability density of two variables X and Y is called joint probability density. In this case, the probability density function of two variables X and Y is called joint probability density function. 

Definition I

Let's consider a pair of continuous variables such as the birthdays of two people. The probability that the pair (X,Y) falls in a certain region D is given by:

Definition II

The variables X and Y are said to be independent random variables if their joint density function is equal to the product of their individual density function.

Example

Solution

 

 

 

Improper integrals on an unbounded region

 We previously evaluated a double integral on a bounded region. What if the region is unbounded? The following theorem comes handy for our rescue.

Theorem

Example

Solution

Practice

Improper Double Integrals

 Definition of an improper integral

It's preferable to deal with improper integrals of functions over rectangles or simple regions where these functions have finitely many discontinuities. However, not such improper integrals can be evaluated. A form of Fubini's theorem allows to evaluate some types of improper double integrals.

Fubini's theorem for improper integrals

Two conditions are necessary for the theorem to work. The function has to be nonnegative on D and has many finitely discontinuities inside D.

Example

Solution

Let's start by plotting the region. 

The function f is continuous on all points of the region D except in (0,0). If you keep the expression of the region, it would be difficult to calculate the double integral. However, if we express the region in 

the following way, the double integral becomes easy to calculate.

Practice

Consider the function f(x,y) = sin(y) /y over the region:

Finding the average value of a function with 2 variables

 We have seen previously that double integrals can be used to calculate volumes and areas. We can also use double integrals to find the average value of a function with 2 variables. 

Definition

If a function f(x,y) is integrable over a plane bounded region D with positive area A(D), then the average value of f is given by:

Note that the area is:

Example

Find the average value of the function f(x,y) = 7xy² on the region bounded by the line y = x and the curve x = ⎷y.

Solution

The figure above represents the region bounded by the line y = x and the curve x =⎷ y

First find the area A(D) where the region D is given by the figure above. We have:

Then the average value of this function over the given region is:

  Practice                                                                                                                                                         

                                                                                                                                                     

 

Calculating areas using double integrals

 Objectives:

1) Finding the area of a rectangular region

2) Finding the area of a nonrectangular region

We can use double integrals to find the area of a rectangular region and a nonrectangular region

Area of a rectangular region

The area of a single rectangular region can be found by a simple geometric formula. It can also be found using double integrals.

By definition, the area of a plane bounded region D is given by:

Area of a nonrectangular region

We have seen that an area can be calculated using a single integral. It can also be calculated using double integrals.

Example

Find the area of the region bounded below by the curve y = x² and above by the line y = 2x in the first quadrant.

Solution

Practice

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Using double integrals to calculate the volume of a solid over general regions

 In a previous post we define the notion of general regions that allows us to calculate doubles integrals over these types of regions. This enables also to calculate the volume of a solid over these regions. Let's dive right into the solution of a problem.

Example

Solution

The solid is limited by the plane x = 0 or yz. the plane  y = 0 or xz, the plane z = 0 or xy and the plane 2x +3y+z = 6. First  we start by drawing the axes of coordinates then we draw part of the plane 2x+3y+z = 0. In order to do the latter, let's determine the coordinates intercepts.

Let's connect the points to form a triangle:

The triangle represents part of the plane 2x+3y+z = 6.. The plane intercepts the planes x = 0, y = 0, z = 0. This gives us the other faces of the tetrahedron, The base is limited by the x axis, the y axis and the line 2x + 3y = 6 as seen on the right. 

Looking at the figure we see a solid with 4 triangular faces, It's called a tetrahedron or triangular pyramid. It consists of the three coordinates planes, the plane z = -2x-3y+6 with the bases limited by the lines y =o, x = 0 and the line 2x +3y = 6.

Practice