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Surface integral of scalar function 
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#1
Apr1510, 12:34 PM

P: 52

1. The problem statement, all variables and given/known data
Find the mass of a spherical surface S of radius R such that at each point (x, y, z) in S the mass density is equal to the distance of (x, y, z) to some fixed point (x_0, y_0, z_0) in S. 2. Relevant equations Integral of a scalar function over a surface. 3. The attempt at a solution I was thinking about converting this into spherical coordinates, but I see no way of doing that nicely since the distance formula would get very messy. I am also assuming they are using the euclidean distance, since this is an intro multivariable course. I don't need help evaluating, just with getting it set up. This is from Vector Calculus, 5e. by Marsden and Tromba, 7.5 #9. 


#2
Apr1510, 01:12 PM

P: 352

It doesn't matter what the fixed point [tex](x_0, y_0, z_0)[/tex] is, so you can make a choice that makes spherical coordinates less horrible.



#3
Apr1510, 01:58 PM

P: 52

[tex]\sqrt{(2Rsin(\phi)cos(\theta))^2 + (2Rsin(\phi)sin(\theta))^2} + (2Rcos(\phi))^2}[/tex] = 2R. So, [tex]\int_0^{2\pi} \int _0^\pi 2R*R^2sin(\phi) \, d\phi d\theta ?[/tex] [tex] = 8\pi*R^3 [/tex] Edit: The answer is supposed to be (16/3)Pi*R^3, so I lost a factor of 2/3 somewhere... 


#4
Apr1510, 02:47 PM

P: 352

Surface integral of scalar function
Well, you probably shouldn't choose [tex](x_0, y_0, z_0)[/tex] to be the origin if your sphere [tex]S[/tex] is centered at the origin. Read the question again; [tex](x_0, y_0, z_0)[/tex] is supposed to lie on [tex]S[/tex].



#5
Apr1510, 03:13 PM

HW Helper
P: 1,495

Make life easy for yourself and take [itex]\theta_0=\phi_0=0[/itex].



#6
Apr1510, 03:21 PM

P: 52




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