Evaluating a double integral over a type II region

 We continue in the evaluation of double integrals over non-rectangular regions. In the previous  post, we solve a double integral over a type I region. In this post, we evaluate the double integral of a function over a type II region.

Example

Solution

Practice

Double integrals over non-rectangular regions

 Goal: Evaluate a double integral over a nonrectangular region

In the previous post, we stated that, in order to evaluate a double integral over a general, non-rectangular region we need to express this region as a type I or type II, The following theorems allow to reach this goal.

Theorem I

Double integrals over nonrectangular regions

Suppose g(x,y) is the extension to the rectangle R of the integrable function f(x,y) defined on the region D where D is inside R. Then g(x,y) is integrable and we define the double integral f(x,y) over D by:

Theorem II

Example

Solution

General regions of Integration

 Objective: recognize when a function f(x,y) is integrable over a general region.

General regions of integration 

Let's consider the following figure:

Here the bounded region D is enclosed by a rectangular region in the plane. Let's suppose a function z = f(x,y) defined over the general planar region D. In order to define integrals of f over the region D, we need to extend the definition of f to include all the points of the rectangular region R. In order to do so, we introduce a new function g defined as follow:

We assume that the region D is a piecewise smooth and continuous function. The function g should be integrable over the region R. This happens as long as the region D is bounded by simple curves. We consider two types of planar bounded regions as stated in the following definition:

Definition

Example

Solution

Practice

Applications of double integrals: calculating the area of a rectangular region using double integral and the volume of a solid with a particular condition

 Objectives:

1) Calculate the volume of a rectangular region

2) Calculate the volume of a region bounded above by a function f(x,y) with f(x,y) ≥ 0.

Applications of double integrals

As we saw from the definition of double integrals, a double integral is a volume of a region R. In this post, we'll study the volume of a rectangular region and the volume of a region bounded by a function f(x,y) provided that f(x,y) ≥ 0.

Volume of a rectangular region

If the region is rectangular, we just calculate the double integral of the constant function f(x,y) = 1.

Definition

The area of the region R is given by:

Example

Solution

Volume of a region bounded above by a function  f(x,y) ≥ 0.

Example

Solution

Practice

Evaluate an iterated integral in two ways

 Goal: Evaluate an iterated integral in two ways

Remember the Fubini's theorem concerning the double integral of a function of two variables f(x,y) that is continuous over a rectangular region R:

Then the double integral is an iterated that can be expressed as:

This means that a double integral can be integrated in two ways. First starting to integrate with respect to y and then x. Second, integrate with respect to x then y. In either way, we obtain the same result.

Example

Let's consider the function f(x, y) = 3x²- y over the rectangular region 

Solution

Practice

Using the Fubini's theorem, evaluate the following integral in two ways:

 

Practice

Application of property VI of double integrals

 Property VI of double integrals (reminder)

In the case where the function f(x,y) is expressed as a product of two functions g(x)  of x only and h(y) of y only, then over the region R = {(x,y)/ a ≤ x ≤ b., c≤ y ≤ d}, the double integral of f(x, y) over the region R can be written as:

Example

Solution

It is clear that the given function f(x,y) is a product of a function of x and a function y. Here we have: g(x) = cosx and h(y) =  e^y. By substituting g(x), h(y), a and b in the equality above, we obtain:

 

Practice

a) Use the properties of double integrals and the Fubini's theorem to evaluate:

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