Surface integral of scalar function

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To find the mass of a spherical surface with radius R and a mass density equal to the distance from a fixed point on the surface, it is suggested to use spherical coordinates. The fixed point can be chosen to simplify calculations, ideally not at the origin if the sphere is centered there. The integral setup leads to a calculation of the mass, initially yielding 8πR^3, but the correct answer is (16/3)πR^3, indicating a miscalculation. The discussion emphasizes the importance of correctly positioning the fixed point to avoid unnecessary complexity in the problem. Ultimately, the participants recognize the need for a simpler approach to reach the correct solution.
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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.
 
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It doesn't matter what the fixed point (x_0, y_0, z_0) is, so you can make a choice that makes spherical coordinates less horrible.
 
ystael said:
It doesn't matter what the fixed point (x_0, y_0, z_0) 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
\sqrt{(2Rsin(\phi)cos(\theta))^2 + (2Rsin(\phi)sin(\theta))^2} + (2Rcos(\phi))^2}
= 2R.

So, \int_0^{2\pi} \int _0^\pi 2R*R^2sin(\phi) \, d\phi d\theta ?

= 8\pi*R^3

Edit: The answer is supposed to be (16/3)Pi*R^3, so I lost a factor of 2/3 somewhere...
 
Last edited:
Well, you probably shouldn't choose (x_0, y_0, z_0) to be the origin if your sphere S is centered at the origin. Read the question again; (x_0, y_0, z_0) is supposed to lie on S.
 
Make life easy for yourself and take \theta_0=\phi_0=0.
 
Cyosis said:
Make life easy for yourself and take \theta_0=\phi_0=0.

Wow, I made that way harder than it had to be. Thank you both, that was driving me nuts!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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