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Area of a Square

  1. May 27, 2014 #1
    So I was reviewing my random process notes. In it there is an integral that they have that I can't seem to get the right derivation of when they try to simply the math for ergodic mean. Basically, you have the following:

    A square from (-T,T) on both the x-axis and y-axis. What they want to do is integrate over the diagonal lines u = x - y where u is a constant.

    So rather than a double integral over dx and dy from (-T, T) for each, they now have one integral du from (-2T, 2T).

    The part that I have a problem with is that the derivation of the area of the diagonal strip u = x-y which the textbook says can be shown to be (2T - u)du for u from 0 to 2T and (2T+u) for u from -2T to 0.

    So the way I did it was to think of that diagonal strip as subtraction of two triangles. Which when I do the math ends up giving me the area as (2T-u)du - 1/2*(du)*(du). I tried to imagine this as the equation for the area of the strip if it were to be turned into a parallelogram, which means that the extra bit of area would be 1/2*(du)*du) which is why its subtracted and as du→0, it reduces to (2T-u)du. But the part where the base of the parallelogram equals (2T-u) just doesn't make sense.

    I am wondering, what is the right way to do this derivation?
     
    Last edited by a moderator: May 27, 2014
  2. jcsd
  3. May 27, 2014 #2

    HallsofIvy

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    No. Since each strip has "infinitesimal" width, think of it as a line, not a parallelogram or
    difference of two triangles. If u> 0, in the upper left half of the square, u is positive while in the lower right half, u is negative.

    If u> 0 (upper left half), one endpoint of the line segment is on y=T and the other on x= -T. Since u= x- y, If y= T, x= u- T so that point is (u-T, T) and if x= -T, u= -T- y, y= -u-T so that point is (-T, -u-T). The length of that line segment is [itex]\sqrt{((u-T)- (-T))^2+ (T- (-u-T))^2}= \sqrt{(u+2T)^2+ (u+2T)^2}= (u+ 2T)\sqrt{2}[/itex]. Taking the thickness to be "du", the area is [itex]\sqrt{2}(u+ 2T)du[/itex].

    If u< 0 (lower right half), one endpoint of the line segment is on x= T and the other is on y= -T. Since u= x- y, if X= T, u= T- y, y= T- u so that point is (T, T- u) and if y= -T, u= x+ T, x= u- T so that point is (u- T, -T). The length of that line segment is [itex]\sqrt{((T- (u-T))^2+ ((T-u)- (-T))^2}= \sqrt{(2T- u)^2+ (2T- u)^2}= (2T- u)\sqrt{2}[/itex]. Taking the width to be "du", the area is [itex]\sqrt{2}(2T- u)du[/itex].
     
    Last edited by a moderator: May 27, 2014
  4. May 27, 2014 #3
    I tried that but was having trouble getting rid of the square root of two term. Do you think thats just an error in the textbook?

    I tried to think of du as [itex]\stackrel{1}{2}sqrt{(dt)^{2}}[/itex], but that only makes it 1/2 rather than root 2.

    The reason I used the parallelogram was that it got rid of the 1/2 term in from and gave me something that was just (2T-u)du
     
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