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Solving Double Integrals Using Symmetry

Date: 11/05/2004 at 15:50:22
From: Darren
Subject: Using symmetry to quick solve double integrals


I'm very curious to learn how symmetry could be used to solve double
integrals, and I've heard many wonderful stories of how quickly the 
method derives results.  I came across a question that asked to solve 
the double integral of z = e^(x^2) - e^(y^2) over the domain [0,1] x 
[0,1], but I'm in need of some guidance on how to apply symmetry 
considerations in solving this question, as the answer is zero!

I can apply symmetry for some straightforward questions.  For example,
I think I understand why the double integral of z = sin(x)cos(.5y)
over the domain [-2,0]X[-2,0] U [0,2]x[0,2] is zero.

My reasoning is:

  1. the domain is symmetric about the y-axis
  2. the function is an odd function of x
  3. thus taking the double integrals over each [-2,0]x[-2,0] and
     [0,2]x[0,2], will cancel the values and equal to zero

But I've no idea how to apply symmetry for the complicated graph of 
e^(x^2) - e^(y^2) or to explain why its double integral over the said 
domain is zero.  Please advise me, I'm curious to learn this
interesting concept.  Thanks!

Date: 11/08/2004 at 19:01:31
From: Doctor Douglas
Subject: Re: Using symmetry to quick solve double integrals !

Hi Darren.

Symmetry allows you to evaluate a quantity whenever one of the 
following happens:

  1.  f(-S) = f(+S)       "f is even".
  2.  g(-S) = -g(+S)      "g is odd".

Since an integral is a sum, the former leads to things adding up to 
twice the half, and the latter leads to things cancelling.  Here are a 
couple of simple 1-dimensional examples:

  Integral[-7,+7] exp(-x^2) dx = 2*Integral[0,+7] exp(-x^2) dx,
  Integral[-7,+7] x*exp(-x^2) dx = 0.

Now, this sometimes happens in 2D as well, when the domain is 
symmetric.  For example, in the original problem above:

  Q = Integral[x,0,1]Integral[y,0,1] exp(x^2) - exp(y^2) dx dy,

we can do the following:  Notice the minus sign in the integrand and
how the two exponential terms are identical in form.  Furthermore, we 
see that under the transformation x<->y, the integrand flips in sign.  
[Remember that symmetry is a property of invariance under a 
transformation].  For every element of area dA = dx*dy that lies
underneath the line y = x (i.e., x<y), there is an element of area
dB = dw*dz, where w = 1 - x and z = 1 - y, and of course z<w.

   +-------+ (1,1)
   | B   . |
   |   .   |
   | .   A |

  Q = Integral[A U B] exp(x^2) - exp(y^2) dx dy
    = Integral[A] exp(x^2) - exp(y^2) dx dy        The domain is
       + Integral[B] exp(x^2) - exp(y^2) dx dy     separable into 
                                                   A and B.

    = Integral[{x,0,1},{y,0,x}] exp(x^2) - exp(y^2) dx dy
       + Integral[{w,0,1},{z,0,w}] exp(z^2) - exp(w^2) dw dz

    = Integral[{x,0,1},{y,0,x}] exp(x^2) - exp(y^2) dx dy
       + Integral[{x,0,1},{y,0,x}] exp(y^2) - exp(x^2) dx dy   

by a substitution of variables in the second term, and we arrive at

    = Integral[A] [{something} - {that same thing}] dx dy   

    = Integral[A] 0 dx dy

    = 0
All of the discussion above with z and w is simply there to make the
statement explicit.  When you are doing this problem, you usually need 
not be so explicit; you can simply say that the integral vanishes by 
symmetry across the line y = x since every contribution to the 
integral from an element of area dA below the line is exactly 
cancelled from a corresponding element of area dB above the line, 
where y>x.

Note that in your other example, z = sin(x)cos(.5y) over the domain
[-2,0]X[-2,0] U [0,2]x[0,2], the domain is NOT symmetric about the
y-axis.  If you flip the domain across the y-axis, the two squares
originally in quadrants I and III move to quadrants II and IV.

              |            quadrant I
              +-------+ (2,2)
           S  | R     |
              |       |
              |       |
      |       |
      |       |            quadrant IV
      |     T | U
  (-2,-2)     |

The key in this example is to realize that the cosine is even and the 
sine is odd (when reflected about the location where their arguments 
are zero), so that the contribution from an area element near the 
letter R is exactly cancelled by the contribution from the area 
element near the letter T.  

   f(T) = f(S)      because cosine(something*y) is even
        = -f(R)     because sine(something else*x) is odd.

Hence you can perform the integration over the square [0,2] x [0,2] in 
the first quadrant, and it will be exactly cancelled by the portion of 
the integral taken over [-2,0] x [-2,0], leaving the final integral 
equal to zero overall.  Note that you need BOTH the fact that the sine 
is odd AND the fact that the cosine is even to justify this 

I hope that this helps clarify things.

- Doctor Douglas, The Math Forum 
Associated Topics:
College Calculus

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