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Double integral computation

  1. Feb 5, 2014 #1
    Can anybody please help me understand the computation of the integral in the attached image. I shall be grateful.
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2014 #2

    HallsofIvy

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    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]
     
  4. Feb 5, 2014 #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.
     

    Attached Files:

    Last edited: Feb 5, 2014
  5. Feb 6, 2014 #4

    HallsofIvy

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    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.

    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.

     
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