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Weirdness from double integrals?

• To: mathgroup at smc.vnet.net
• Subject: [mg95708] Weirdness from double integrals?
• From: Erik Max Francis <max at alcyone.com>
• Date: Sun, 25 Jan 2009 06:53:10 -0500 (EST)

I'm new to Mathematica 7 (fiddling with a trial version) and I've bumped 
into something that's confusing me.

Just to make sure I understand Mathematica, I'm playing with finding the 
gravitational field around extended objects using integration.  So far 
I've had no problem (e.g., lines, rings, planes, even Newton's shell 
theorem popped out easily enough), but I'm getting something that's 
confusing about double definite integration the Integrate that's leaving 
me puzzled.  Here's the (cleaned-up) transcript from math:

In[1]:= p = {0, 0, h}

Out[1]= {0, 0, h}

In[2]:= s = {x, y, 0}

Out[2]= {x, y, 0}

In[3]:= q = p - s

Out[3]= {-x, -y, h}

In[4]:= dm = rho (* dx dy *)

Out[4]= rho

In[5]:= integrand = -G * dm * q/(q.q)^(3/2)

               G rho x            G rho y              G h rho
Out[5]= {-----------------, -----------------, -(-----------------)}
            2    2    2 3/2    2    2    2 3/2      2    2    2 3/2
          (h  + x  + y )     (h  + x  + y )       (h  + x  + y )

In[7]:= Integrate[integrand, {x, -x0, x0}, {y, -y0, y0}]

Out[7]= {0, 0, 0}

In[9]:= Integrate[Integrate[integrand, {x, -x0, x0}, Assumptions -> x0 > 
0 && h > 0], {y, -y0, y0}, Assumptions -> x0 > 0 && y0 > 0 && h > 0]

                                                x0 y0
Out[9]= {0, 0, -2 G rho (Pi - 2 ArcCot[----------------------])}
                                                2     2     2
                                        h Sqrt[h  + x0  + y0 ]

It would seem to me, less assumptions, that Out[7] and Out[9] should be 
the same, but Out[7] is clearly wrong (read: "not what I expected" :-) 
-- while the x and y components of the integration are zero by symmetry, 
the z component clearly should not be.  Out[9] looks correct, but I 
thought these would have been the same computation; converting In[7] to 
traditional form in the front end appears to confirm my understanding 
that it does indeed represent a double definite integral over the integrand.

So what is there about doing double definite integrals with Integrate 
that I'm missing?  Why aren't these computations the same?

Thanks.

-- 
Erik Max Francis && max at alcyone.com && http://www.alcyone.com/max/
  San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
   He who can, does. He who cannot, teaches.
    -- George Bernard Shaw

• Follow-Ups:
  • Re: Weirdness from double integrals?
    • From: danl@wolfram.com