# Surface integral of scalar function

by malicx
Tags: function, integral, scalar, surface
 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.
 P: 352 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.
P: 52
 Quote by ystael 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...

 P: 352 Surface integral of scalar function 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 $$S$$.
 HW Helper P: 1,495 Make life easy for yourself and take $\theta_0=\phi_0=0$.
P: 52
 Quote by Cyosis 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!

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