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Re: Definite integrals in Mathematica
*To*: mathgroup at smc.vnet.net
*Subject*: [mg72257] Re: [mg72145] Definite integrals in Mathematica
*From*: Darren Glosemeyer <darreng at wolfram.com>
*Date*: Sat, 16 Dec 2006 05:18:32 -0500 (EST)
*References*: <200612131140.GAA23742@smc.vnet.net>
As Dimitris pointed out, Integrate gives a correct result. However, I
think the integral you have computed is not the integral of interest. I
think what you want is the integral of f(r) over the volume of the
sphere. Writing this as an integration with respect to r gives the
result from the literature.
In[2]:= dVdr = D[4/3Pi*r^3, r]
Out[2]//InputForm= 4*Pi*r^2
In[3]:= f[r_] = (h^2 - r^2)^3
Out[3]//InputForm= (h^2 - r^2)^3
In[4]:= Integrate[f[r]*dVdr, {r, 0, h}]
Out[4]//InputForm= (64*h^9*Pi)/315
Darren Glosemeyer
Wolfram Research
some guy named Dave wrote:
> Hey all,
>
> Hmm, I tried posting this yesterday, but it hasn't shown up yet. So,
> appologies if this gets double-posted.
>
> I'm a newbie to Mathematica, and I have a function that I want to integrate.
> It's basically a smoothing function over a vector space, and I want to
> normalize it. The function itself looks like this:
>
> f(r) = (h^2 - r^2)^3
>
> where r is the scalar length ||x|| of some displacement vector x, and h is a
> constant. We are really only interested in values of r in [0,h], and for 2D
> vectors. However, I have some research literature involving 3D vectors that I
> want to compare against, just for sanity.
>
> So, if x is a 3D vector, then the surface of the sphere at radius r should be
> (4/3)(f(r))^3. We should then be able to get the definite integral from
> Mathematica by doing something like:
>
> Integrate[ 4/3 * Pi( (h^2 - r^2)^3 )^3, {r, 0, h}]
>
> When I do so, I get back
>
> 262144 pi h^19 / 692835
>
> However, the literature I have says that the integral should yield
>
> 64 pi h^9 / 315
>
> So, have I screwed something up? Is the literature wrong, perhaps?
>
> Thanks.
>
> Dave
>
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