Surface integral of scalar function

  • Thread starter malicx
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  • #1
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Homework Statement


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.


Homework Equations


Integral of a scalar function over a surface.

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.
 

Answers and Replies

  • #2
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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
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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.

So basically, choose (x_0, y_0, z_0) to be 0, and using spherical coordinates, the distance is
[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...
 
Last edited:
  • #4
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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
Cyosis
Homework Helper
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Make life easy for yourself and take [itex]\theta_0=\phi_0=0[/itex].
 
  • #6
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Make life easy for yourself and take [itex]\theta_0=\phi_0=0[/itex].

Wow, I made that way harder than it had to be. Thank you both, that was driving me nuts!
 

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