Understanding the Computation of Double Integrals: Can You Help?

In summary, the conversation is discussing the computation of an integral with a double integral in the attached image. The goal is to express the double integral in terms of a single integral by converting the function c(t_1- t_2) into a function of a single variable. This can be achieved by defining \tau= t_1- t_2 and \sigma= t_1+ t_2. The region in question is a rectangle in the \tau, \sigma plane with its diagonals parallel to the axes. The lines \tau= t_1+ t_2 and \sigma= t_1- t_2 are perpendicular, creating an "orthogonal" coordinate system. The conversation also includes a request
  • #1
inviziblesoul
2
0
Can anybody please help me understand the computation of the integral in the attached image. I shall be grateful.
 

Attachments

  • Problem.jpg
    Problem.jpg
    18.5 KB · Views: 362
Physics news on Phys.org
  • #2
What exactly is the integral you are asking about? The integral on the left is simple, just the integral of [itex]c(t_1- t_2)[/itex] over the square [itex]-T\le t_1\le T[/itex] and [itex]-T\le t_2\le T[/itex]. Is c a constant or a function of [itex]t_1- t_2[/itex]? If a function, then what function?

Perhaps you are trying to convert the function [itex]c(t_1- t_2)[/itex] to a function of a single variable by defining [itex]\tau= t_1- t_2[/itex]? The region is still two dimensional and even though c can be set as a function of one variable, you will need another variable, say [itex]\sigma= t_1+ t_2[/itex]. Then [itex]t_1= (\tau+ \sigma)/2[/itex] and [itex]t_2= (\sigma- \tau)/2[/itex] so that the boundaries of the square become [itex]t_1= (\tau+ \sigma)/2= T[/itex] so [itex]\tau+ \sigma= 2T[/itex], [itex]t_1= (\tau+ \sigma)/2= -T[/itex] so [itex]\tau+ \sigma= -2T[/itex], [itex]t_2= (-\tau+ \sigma)/2= T[/itex] so [itex]-\tau+ \sigma= 2T[/itex], and [itex]t_1= (-\tau+ \sigma)/2= -T[/itex] so [itex]\tau- \sigma= 2T[/itex].

That is a rectangle in the [itex]\tau[/itex], [itex]\sigma[/itex] plane with its digonals parallel to the axes. Overall, [itex]\tau[/itex] goes from -2T to 2T but because the upper and lower boundaries change formula at [itex]\tau= 0[/itex] you should do it as two separate integrals. For [itex]\tau= -2T[/itex] to [itex]\tau= 0[/itex], the lower boundary is [itex]\tau+ \sigma= 2T[/itex] or [itex]\sigma= 2T- \tau[/itex] and the upper boundary is [itex]-\sigma+ \tau= 2T[/itex] or [itex]\sigma= \tau- 2T[/itex]. For [itex]\tau= 0[/itex] to [itex]\tau= 2T[/itex], the lower boundary is [itex]=-\sigma+ \tau= -2T[/itex] or [itex]\sigma= -\tau- 2T[/itex] and the upper boundary is [itex]sigma+ \tau= 2T[/itex] or [itex]\sigma= -\tau+ 2T[/itex].

So the integral can be written
[tex]\int_{-2T}^0\int_{2T-\tau}^{\tau+ 2T} c(\tau)d\sigma d\tau+ \int_0^{2T}\int_{-\tau- 2T}^{\tau- 2T} c(\tau) d\sigma d\tau[/tex]
 
  • #3
Thank you very much for your excellent efforts and this great explanation. However, I am not clear at certain points.

You have rightly pointed out: the aim here is to express the double integral in terms of a single integral. Furthermore, C is a function of the difference [itex]\tau=t_1−t_2[/itex].

I did not understand your phrase <<That is a rectangle in the τ, σ plane with its digonals parallel to the axes.>> How do you know that its a rectangle and its diagonals are parallel to the axes (the [itex]\tau, \sigma[/itex] axes ?).
and how did you choose [itex]\sigma = t_1 + t_2?[/itex] why not some other function?
I will greatly appreciate if you can kindly refer me some reading on this topic.

I have attached my solution as well. I have not introduced a new variable, however, I have used [itex]\tau[/itex] and [itex]t_1[/itex].

Thank you for your time.
 

Attachments

  • DOC050214-05022014164355.pdf
    338.3 KB · Views: 199
Last edited:
  • #4
inviziblesoul said:
Thank you very much for your excellent efforts and this great explanation. However, I am not clear at certain points.

You have rightly pointed out: the aim here is to express the double integral in terms of a single integral. Furthermore, C is a function of the difference [itex]\tau=t_1−t_2[/itex].

I did not understand your phrase <<That is a rectangle in the τ, σ plane with its digonals parallel to the axes.>> How do you know that its a rectangle and its diagonals are parallel to the axes (the [itex]\tau, \sigma[/itex] axes ?).
The "coordinate axes" in a [itex]\tau[/itex], [itex]\sigma[/itex] coordinate system are the lines [itex]\tau= 0[/itex] and [itex]\sigma= 0[/itex] which mean [itex]t_1- t_2= 0[/itex] and lines parallel to that.

and how did you choose [itex]\sigma = t_1 + t_2?[/itex] why not some other function?
I will greatly appreciate if you can kindly refer me some reading on this topic.
The lines [itex]\tau= t_1+ t_2= constant[/itex] is the same as [itex]t_2= -t_1+ constant[/itex] have slope -1. The lines [itex]\sigma= t_1- t_2= constant[/itex] or [itex]t_2= t_1- constant[/itex] have slope 1. They are perpendicular so we still have an "orthogonal" coordinate system.

I have attached my solution as well. I have not introduced a new variable, however, I have used [itex]\tau[/itex] and [itex]t_1[/itex].

Thank you for your time.
 

1. What is a double integral?

A double integral is a type of mathematical calculation that involves evaluating a function over a two-dimensional region. It can be thought of as finding the volume under a surface in three-dimensional space.

2. How is a double integral computed?

A double integral is typically computed by first determining the limits of integration for both the x and y variables, then breaking the region into small, rectangular elements. The function is then evaluated at each element and the results are summed together to find the total volume.

3. What is the purpose of using a double integral?

Double integrals are used in many areas of science and engineering to calculate properties such as area, volume, and mass. They are particularly useful in physics and engineering for calculating the work done by a force over a given region.

4. What is the difference between a single and a double integral?

A single integral computes the area under a curve in one dimension, while a double integral computes the volume under a surface in two dimensions. Essentially, a double integral involves performing a single integral over a range of values for a second variable.

5. Can a double integral be applied to any function?

Yes, a double integral can be applied to any continuous function over a two-dimensional region. However, the computation may become more complex for certain functions, and may require advanced techniques such as change of variables or integration by parts.

Similar threads

Replies
1
Views
793
  • Calculus
Replies
11
Views
2K
Replies
10
Views
408
Replies
1
Views
834
Replies
4
Views
1K
  • Calculus
Replies
5
Views
1K
Replies
31
Views
753
Replies
3
Views
535
  • Calculus
Replies
1
Views
935
  • Calculus
Replies
24
Views
3K
Back
Top