Surface integral on a 3D region (Solution)

*To*: mathgroup at smc.vnet.net*Subject*: [mg111565] Surface integral on a 3D region (Solution)*From*: dr DanW <dmaxwarren at gmail.com>*Date*: Thu, 5 Aug 2010 07:01:07 -0400 (EDT)

Here is the solution I worked out for the surface integral question I posed to the group about a week ago. I hope someone else can make use of this solution in the future. These are the expressions that, when equal to zero, define the surfaces of my solid region, which is a circular cylinder with one flat end and one slanted end. In[1]:= L = 15; R = 3.5; \[Theta] = 40.*Degree; In[2]:= cylinder = -R^2 + $y^2 + $z^2; piston = $x; termination = $z*Cos[\[Theta]] + ($x - L)*Sin[\[Theta]]; In[5]:= xmax = L + R/Tan[\[Theta]]; So, the solid region is defined as In[6]:= tubeRegion = cylinder <= 0 && piston >= 0 && termination <= 0 Out[6]= -12.25 + $y^2 + $z^2 <= 0 && $x >= 0 && 0.642788 (-15 + $x) + 0.766044 $z <= 0 My problem is that I want to be able to integrate a function over the entire surface of this region. That was not covered in my calculus class, since I have boundaries that are not on constant coordinate planes. The first thing to do is to break the problem down to integrating over each of the three surfaces that bound the region. To work one surface, my first thought was to change one of the inequalities to an equality and do a 3D NIntegrate, but that gave me zero since the resulting region has zero volume (duh.) So, it was evident that I had to reduce the problem to 2D. So, pick a variable and solve the surface equation for that variable: In[7]:= xform = Solve[cylinder == 0, $y] Out[7]= {{$y -> -0.5 Sqrt[49. - 4. $z^2]}, {$y -> 0.5 Sqrt[49. - 4. $z^2]}} I will use this variable transformation to eliminate the $y variable from the remaining expressions, essentially turning the 3D problem into 2D problem in the parameters $x and $z. Looking up some formulas for parametric surface integration in MathWorld, I calculate the element of surface in the new coordinates In[8]:= ds = (Norm[Cross[D[#1, $x], D[#1, $z]]] & ) /@ ({$x, $y, $z} /. xform) Out[8]= {Sqrt[1. + 4. Abs[$z/Sqrt[49. - 4. $z^2]]^2], Sqrt[ 1. + 4. Abs[$z/Sqrt[49. - 4. $z^2]]^2]} I also use the transformation of the remaining two bounding inequalities. In this case, I cheat since one of the bounding conditions is trivial and I can put them in directly as limits of integration. In[9]:= select = Boole /@ (termination <= 0 /. xform) Out[9]= {Boole[0.642788 (-15 + $x) + 0.766044 $z <= 0], Boole[0.642788 (-15 + $x) + 0.766044 $z <= 0]} To integrate a function f[x,y,z] over the one surface we have been working with, the integral would be (* NIntegrate[(f[x,y,z]/.xform).(ds select),{$x,0,xmax}, {$z,-3.35,3.35}] *) For example, for the trivial case where we only want the surface area, the function f[x,y,z] = 1, so the integral evaluates In[11]:= NIntegrate[(1 /. xform) . (ds*select), {$x, 0, xmax}, {$z, -3.35, 3.35}] Out[11]= 268.164 Now repeat for the other two surfaces. Other solutions were suggested, such as giving the surfaces a small thickness, performing a volume integration, and dividing by the thickness, or using Dirac functions to define the surfaces. I have not tried either of these. Thanks to everybody for their help. Enjoy, Daniel

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