Finding a volume of a solid in polar coordinates using double integrals

 We continue with solving problems of volume of solids in polar coordinates using double integrals. Here is another example.

Example 2

Solution

Let's first express the equation of the region which is a disk in polar coordinates. We need to find θ and r.

The equation of the disk can be expressed as::

This is a set of circles with radiuses less than or equal to1. These circles are centered in (1,0). The center can be found by rewriting the left part of the above inequality as (x-1)² + (y-0)².  The angle θ varies from 0 to 2ℼ..

Expanding the square term in the equation (x-1)² + y² = 1, we have: 

By simplification, we get : 

By substituting x = cosθ and y = sinθ, we have:

Solving this equation, we find: 

The disk on the xy plane can be expressed on the following region as:

Let's express z in polar coordinates by substituting x and y in the equation of z. We obtain z = 4-r².

Let's calculate the volume. We have:

Let's calculate the inner integral:

=  

= 8cos²θ−4cos⁴θ

Let's integrate with respect to the outer integral:

Calculating this expression we have:

V = 5/2.2π +5/2sin4π-1/8sin8π- 0 = 5𝛑/2

Volume of a solid in polar coordinates using double integrals

 As in rectangular coordinates, if a solid S is bounded by the surfaces z = f(r, θ) and the surfaces r = a, r = b, θ = ɑ, θ = β, then the volume V of S can be found by integration using the formula:

If the base of the solid can be described as D = {(r, θ), ɑ ≤ θ ≤ β h₁(θ) ≤ θ ≤ h₂(θ)}, the volume V becomes:

Example

Solution

By the method of double integration, the volume is the iterated integral of the form:

Evaluating a Double Integral over General Polar Regions of Integration

 Goal: Evaluate a double integral over a general polar of integration

To evaluate the double integral of a continuous function over polar general regions using iterated integrals, we consider the types I and II regions used previously in the calculation of double integrals over general regions in rectangular coordinates. We write polar equations as r = f(θ) rather than θ = f(r). The general polar region is defined by:

 

 

The figure above represents the general polar region between ɑ ≤θ ≤ β and h₁ (θ) ≤θ ≤ h₂(θ)

Theorem

Example

Solution

Practice

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