Application of property V of double integrals

 Objective: apply the fifth property of the double integrals

Example:

Solution

The fifth property of the integral states:

This inequality represents also:

In the given inequality m = 2 and M = 13. Let's substitute them and f(x,y) in the inequality above:

Let's calculate the integrals of the constants 2 and 13 as well as that of the function f(x,y). This leads to write the following inequality:

 

Let's calculate the integral of the left and that of the right and then substitute:

Practice

Properties of double integrals

 Goal: Recognize and use some of the properties of double integrals

The properties of double integrals are very helpful to compute them. Here I list their six properties, Properties 1 and 2 are referred as the linearity of the integral. Property 3 is the additivity of the integral. Property 4 is the monotonicity of the integral, and property 5 is used to find the bounds of the integral. Property6 is used if f(x,y) is a product of two functions g(x) and h(y).  

Example

Solution

Let's apply properties I and II: 

Let's convert the double integrals to iterate:

Integrate with respect to x holding y;

Integrate with respect to y:

Practice

Calculate the following double integral 

over the rectangular region R given by 0 ≤ x ≤ 3., 0≤ y≤2

Double integrals of a function f(x, y) over a rectangular region R (continued)

 Objectives:

1. Find a notation for the double integral of a function f(x,y) over a rectangular region R

2. Evaluate a double integral over a rectangular region by writing it as an iterated integral.

Notation.

Remember that the double integral of a function f(x,y) over a rectangular region R is given by:

Iterated Integrals.

The process of calculating double integrals can be lengthy especially if m and n become larger numbers. Therefore it's better to find a way to calculate the integrals without using limits and double sums. The method used consists in breaking the double integral in simple integrals where one integral is evaluated to one variable and the other integral to the other variable, This process is called iterated integral.

The fact that double integrals can be expressed as iterated integral is expressed in Fubini's theorem.

Integrating first with respect to y and then to x to find the area A(x) and then the volume V.

Integrating first with respect to x and then to y to find the area A(y) and then the volume V.

Example

Solution

Since dx = dy, the Fubini's theorem allows to use [a b}or [c d] as upper limit of integration for a function f(x,y) defined over a rectangular region R = [a b] . [c d]. To calculate the double integral, we transform it in iterated integrals.

 

Double integral of a function f(x,y) over a rectangular region R

 Objective: Recognize when a function of two variables is integrable over a rectangular region

Double integral of a function f(x,y) over a rectangular region R

Let's start by considering the space above a rectangular region R. Let z = f(x,y) ≥ 0 be a function of two variables defined over R as follows:

The graph of f is a surface above the xy plane where z = f(x,y) is the height of the surface at the point (x,y). Let's consider the volume of the solid between the surface S and the rectangular region R. The base of the solid is the rectangle R. Our goal is to find the volume V of the solid S.

Let's divide the interval [a b] in m sub-intervals and the interval [c d] in n sub-intervals. This allows to divide the rectangular region R into small rectangles Rij with area ΔA and sides Δx and Δy. Therefore we have:

Let's consider a thin rectangular box above Rij with height f(xij, yij).

Considering all the thin rectangular boxes for all the subrectangles, we obtain an approximate volume of the solid S as follows:

This sum is known as the double Rieman sum and can be used to approximate the volume of the solid. The double sum means that, for each subrectangle, we evaluate the function at the chosen point, multiply by the area  and add all the results.

As seen in the case of a single variable function, we obtain a better approximation when m and n become larger.

Here we are ready to define the double integral of a function f(x,y) over a rectangular region R in the xy plane.

Definition

The double integral of the function f (x,y) over the rectangular region R in the xy plane is given by:

Example

Solution

Since the number of subintervals respectively on the x-axis and the y-axis is known i.e  m = n =2, we don't use the limit notation to calculate the volume. Therefore:

Then V = [3(1)² - 1] * 1 + [3(2)² - 1] * 1 + [3(1)² - 2] * 1 + [3(2)² - 2] * 1

V = (3 - 1) * 1 + (12 - 1) * 1 + (3 - 2) * 1 + (12 - 2) * 1

V =  2 + 11 + 1 + 10

V = 24

Then V =  [3(1/2)² - 1/2] * 1 + [3(3/2)² - 1/2]  * 1 + [3(1/2)² - 3/2] * 1 + [3(3/2)² - 3/2] *1

V = (3/4 - 1/2) + (27/4 - 1/2) + (3/4 - 3/2) + (27/4 - 3/2)

V = 3/4 - 2/4 + 27/4 - 2/4 + 3/4 - 6/4

V = 44/4 = 11.

Practice

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Review of functions of more than 2 variables

In the previous post we reviewed the function of two variables. In this post we will review certain notions regarding the functions of more than 2 variables:

Functions of more than two variables

Level surface of a function of three variables

Limit of a function of three or more variables

Partial derivatives of a function of three variables

Review of a function of two variables: Maxima and Minima

 Maxima and Minima Problems

In this post we review the following notions:

Critical points in a function of three variables

Local extremum and the saddle point in a function of two variables

Absolute maxima and minima

Review of functions of two variables: Directional derivative and the gradient

 Directional derivatives and the gradient

Review the following posts:

Directional derivatives of a function of two variables

Directional derivatives of a function of two variables (continued) 

Maximum directional derivative

The next post will be about Critical points, local extremum, absolute maxima and minima