Re: Weirdness from double integrals?

*To*: mathgroup at smc.vnet.net*Subject*: [mg95744] Re: [mg95708] Weirdness from double integrals?*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Sun, 25 Jan 2009 21:51:46 -0500 (EST)*Reply-to*: hanlonr at cox.net

p = {0, 0, h}; s = {x, y, 0}; q = p - s; dm = rho; integrand = -G*dm*q/(q.q)^(3/2); They give identical results when both have the same assumptions Integrate[integrand, {x, -x0, x0}, {y, -y0, y0}, Assumptions -> x0 > 0 && y0 > 0 && h > 0] === Integrate[Integrate[integrand, {x, -x0, x0}, Assumptions -> x0 > 0 && h > 0], {y, -y0, y0}, Assumptions -> x0 > 0 && y0 > 0 && h > 0] True Bob Hanlon ---- Erik Max Francis <max at alcyone.com> wrote: ============= 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

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**Re: Weirdness from double integrals?**

**Re: Weirdness from double integrals?**

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