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Triple integration w/spherical coordinates

  • Thread starter MAins
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  • #1
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1.

"Find the mass of part of the solid sphere x^2 + y^2 + z^ 2 ≤ 25 in the 1st octant x ≥ 0, y ≥ 0, z ≥ 0 where mass density is f (x, y, z ) = (x^2 + y^2 + z^2 )^3/2 ."

3.

These problems are really stumping me! I need somebody to work it out/explain it to me! What will the limits of integration be for the following question? What do i integrate? I know I need to transform it to spherical coordinates... but beyond that I'm lost.

I know it's a triple integral:
m = ∫∫∫(x^2 + y^2 + z^2)^3/2 dzdydx
transforming to spherical co-ordinates:
0 ≤ rho ≤ 5
0 ≤ theta ≤ ??? (how do I figure this out?)
0 ≤ phi ≤ ????? (ditto)

dzdydx = rho^2 sinphi drho dphi dtheta
What does f (x, y,z) transform to and how do I figure it out?
 

Answers and Replies

  • #2
Dick
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x^2+y^2+z^2=rho^2. What does that make f(x,y,z)? phi is the polar angle and theta is the equatorial angle. What range of these keeps you in the first octant? Refer to a picture of your preferred system of spherical coordinates.
 
  • #3
tiny-tim
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… keep it simple …

Hi MAins! :smile:

You're making this too complicated:cry:

It's one-eight of the mass of the whole sphere.

Divide it into spherical shells, of radius r, from r = 0 to 5.

Then the mass of each shell is … ? :smile:
 
  • #4
HallsofIvy
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But just in case, you would like to learn how to do these problems!

Since [itex]\rho= \sqrt{x^2+ y^2+ z^2}[/itex], [itex](x^2+ y^2+ z^2)^{3/2}= \rho^3[/itex]. Thats the function you want to integrate.

Yes, you are correct that the differential of volume in spherical coordinates is [itex]\rho sin^2(\phi)d\rho d\theta d\phi[/itex]

Now, to determine what the limits of integration should be, think about what the variables in spherical coordinates mean. [itex]\rho[/itex] measures the distance from the origin, (0,0,0) to a point. Since your sphere is centered at (0,0,0) and has radius 5, [itex]\rho[/itex] must go from 0 to 5. [itex]\theta[/itex] measures the angle around the "equator" (think of it as "longitude"). For the full sphere, it goes from 0 to [itex]2\pi[/itex]. Here, you have just 1/4 of a full circle: so [itex]\theta[/itex] goes from 0 to ?? [itex]\phi[/itex] measures the angle down from the z-axis to the point (think of it as "co-latitude". For the full sphere, it goes from 0 to [itex]\pi[/itex]. You only want to go from the z-axis to the xy-plane, 1/2 way: so [itex]\phi[/itex] goes from 0 to ??
 

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