Proving Properties of Double Integrals

I will also follow your advice and organize the lemmas at the top. Thanks!In summary, the property states that if m and M are real numbers and f(x,y) is between them for all (x,y) in D, then the double integral over D of f(x,y) is between m times the area of D and M times the area of D. This can be proven by using the given information and some known properties of integrals.
  • #1
_N3WTON_
351
3

Homework Statement


Prove the following property:
If [itex] m <= f(x,y) <= M \hspace{2 mm} \forall (x,y) \in D,[/itex] then:
[itex] mA(D) <= \int\int f(x,y)\,dA <= MA(D) [/itex]

Homework Equations


I use a few other known properties in the proof (see below)

The Attempt at a Solution


First, I should state that this problem is for Multivariable Calc. My ability to do proof's is severely lacking, and it is a skill I would like to learn (although I have yet to take a course on it, perhaps next semester). So I am hoping you all will be as critical as possible so I can learn from my mistakes. Also, I am not sure if I am using correct terminology, so please inform me if I am not. This is my attempt:
Given that m and M are real numbers such that:
[itex] m <= f(x,y) <=M \hspace{2 mm}\forall (x,y) \in D [/itex]
It is known that:
[itex] \int\int mdA <= \int\int f(x,y) \,dA <= \int\int MdA [/itex]
Lemma:
[itex] {if} \hspace{2 mm} f(x,y)>=g(x,y) \hspace{2 mm}\forall (x,y) \in D [/itex]
then:
[itex] \int\int f(x,y)\,dA >= \int\int g(x,y)\,dA [/itex]
Lemma:
[itex] \int\int cf(x,y)\,dA = c\int\int f(x,y)\,dA [/itex]
then:
[itex] m\int\int\,dA <= \int\int f(x,y)\,dA <= M\int\int\,dA [/itex]
Lemma:
[itex] \int\int\,dA = A(D) [/itex]
Therefore:
[itex] mA(D) <= \int\int f(x,y)\,dA <= MA(D) [/itex]
Proved.
Again, please be very critical, I want to learn how I can become better at this. :)
 
Last edited:
Physics news on Phys.org
  • #2
_N3WTON_ said:

Homework Statement


Prove the following property:
If [itex] m <= f(x,y) <= M \hspace{2 mm} \forall (x,y) \in D,[/itex] then:
[itex] mA(D) <= \int\int f(x,y)\,dA <= MA(D) [/itex]
What are m(A) and M(A)?
_N3WTON_ said:

Homework Equations


I use a few other known properties in the proof (see below)

The Attempt at a Solution


First, I should state that this problem is for Multivariable Calc. My ability to do proof's is severely lacking, and it is a skill I would like to learn (although I have yet to take a course on it, perhaps next semester). So I am hoping you all will be as critical as possible so I can learn from my mistakes. Also, I am not sure if I am using correct terminology, so please inform me if I am not. This is my attempt:
Given that m and M are real numbers such that:
[itex] m <= f(x,y) <=M \hspace{2 mm}\forall (x,y) \in D [/itex]
It is known that:
[itex] \int\int mDA <= \int\int f(x,y) \,dA <= \int\int MdA [/itex]
Lemma:
[itex] {if} \hspace{2 mm} f(x,y)>=g(x,y) \hspace{2 mm}\forall (x,y) \in D [/itex]
then:
[itex] \int\int f(x,y)\,dA >= \int\int g(x,y)\,dA [/itex]
Lemma:
[itex] \int\int cf(x,y)\,dA = c\int\int f(x,y)\,dA [/itex]
then:
[itex] m\int\int\,dA <= \int\int f(x,y)\,dA <= M\int\int\,dA [/itex]
Lemma:
[itex] \int\int\,dA = A(D) [/itex]
Therefore:
[itex] mA(D) <= \int\int f(x,y)\,dA <= MA(D) [/itex]
Proved.
Again, please be very critical, I want to learn how I can become better at this. :)
 
  • #3
Mark44 said:
What are m(A) and M(A)?
m and M are real numbers...A(D) is the area of some closed, bounded region in the xy-plane
 
  • #4
_N3WTON_ said:

Homework Statement


Prove the following property:
If [itex] m <= f(x,y) <= M \hspace{2 mm} \forall (x,y) \in D,[/itex] then:
[itex] mA(D) <= \int\int f(x,y)\,dA <= MA(D) [/itex]

Homework Equations


I use a few other known properties in the proof (see below)

The Attempt at a Solution


First, I should state that this problem is for Multivariable Calc. My ability to do proof's is severely lacking, and it is a skill I would like to learn (although I have yet to take a course on it, perhaps next semester). So I am hoping you all will be as critical as possible so I can learn from my mistakes. Also, I am not sure if I am using correct terminology, so please inform me if I am not. This is my attempt:
Given that m and M are real numbers such that:
[itex] m <= f(x,y) <=M \hspace{2 mm}\forall (x,y) \in D [/itex]
It is known that:
[itex] \int\int mdA <= \int\int f(x,y) \,dA <= \int\int MdA [/itex]
Is the above given information? If not, how is it "known that ..."? You have cited the lemmas below, so I can see where they came from. I don't doubt that it's true, though.
_N3WTON said:
Lemma:
[itex] {if} \hspace{2 mm} f(x,y)>=g(x,y) \hspace{2 mm}\forall (x,y) \in D [/itex]
then:
[itex] \int\int f(x,y)\,dA >= \int\int g(x,y)\,dA [/itex]
Lemma:
[itex] \int\int cf(x,y)\,dA = c\int\int f(x,y)\,dA [/itex]
then:
[itex] m\int\int\,dA <= \int\int f(x,y)\,dA <= M\int\int\,dA [/itex]
Lemma:
[itex] \int\int\,dA = A(D) [/itex]
Therefore:
[itex] mA(D) <= \int\int f(x,y)\,dA <= MA(D) [/itex]
Proved.
Again, please be very critical, I want to learn how I can become better at this. :)
I would organize things differently. Put the lemmas at the top so that they don't interrupt the logic of your argument.
It's given that m <= f(x, y) <= M, for all (x, y) in D,
so ##\int\int mdA \leq \int\int f(x, y) dA \leq \int\int MdA##,
hence ##m \int\int dA \leq \int\int f(x, y) dA \leq M\int\int dA##,
and the conclusion follows from this.
 
  • #5
Mark44 said:
Is the above given information? If not, how is it "known that ..."? You have cited the lemmas below, so I can see where they came from. I don't doubt that it's true, though.
The above is not given information, thanks for pointing that out. So maybe it would work better if I stated that I am "taking the double integral over the entire region D" instead of saying "it is known"?
 
  • #6
Mark44 said:
I would organize things differently. Put the lemmas at the top so that they don't interrupt the logic of your argument.
It's given that m <= f(x, y) <= M, for all (x, y) in D,
so ##\int\int mdA \leq \int\int f(x, y) dA \leq \int\int MdA##,
hence ##m \int\int dA \leq \int\int f(x, y) dA \leq M\int\int dA##,
and the conclusion follows from this.
Thanks for the advice! :)
 
  • #7
_N3WTON_ said:
The above is not given information, thanks for pointing that out. So maybe it would work better if I stated that I am "taking the double integral over the entire region D" instead of saying "it is known"?
I think that's better.
 
  • #8
Mark44 said:
I think that's better.
Awesome, thanks for the advice!
 

1. What is a double integral and how is it different from a single integral?

A double integral is a mathematical tool used to calculate the area under a two-dimensional surface. It is different from a single integral because it takes into account both the width and height of a surface, whereas a single integral only considers one variable.

2. How do you prove properties of double integrals?

The properties of double integrals can be proven using mathematical techniques such as substitution, integration by parts, and trigonometric identities. These techniques allow you to manipulate the integrand to a more manageable form, making it easier to evaluate the integral and prove its properties.

3. What are some common properties of double integrals?

Some common properties of double integrals include linearity, symmetry, and the change of variables property. Linearity states that the integral of a linear combination of functions is equal to the linear combination of their integrals. Symmetry refers to the fact that the double integral of an odd function over a symmetric region is equal to zero. The change of variables property allows for the substitution of variables in the integrand to simplify the double integral.

4. Can double integrals be used to calculate volumes and surface areas?

Yes, double integrals can be used to calculate volumes and surface areas. This is done by integrating over a three-dimensional region or surface, respectively. The double integral can be thought of as a sum of infinitesimally small volumes or surface areas, which when added together, give the total volume or surface area.

5. What are some real-world applications of double integrals?

Double integrals have various real-world applications, such as in physics, engineering, and economics. They can be used to calculate the center of mass of a two-dimensional object, the work done by a force, or the total cost of production in a manufacturing process. They are also used in image processing and computer graphics to determine the brightness and color of pixels in an image.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
474
  • Calculus and Beyond Homework Help
Replies
6
Views
489
  • Calculus and Beyond Homework Help
Replies
3
Views
495
  • Calculus and Beyond Homework Help
Replies
20
Views
350
  • Calculus and Beyond Homework Help
Replies
4
Views
796
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
809
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top