Plane intersecting a sphere

  • #1
The unit sphere x2 + y2 + z2 =1 with density =1 is cut by the plane z=1/2. Find the volume and centroid of each piece.

As of now, i have (for the top piece) the integral of [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] r2 Sin[p] dr dp dt, with theta from 0 to 2 pi, phi from 0 to pi/2, and row from 1/(2Cos[phi]) to 1, but when i multiply in rCos[p] to find the z coordinate of the centroid, I get 3/8 which is below that part of the sphere.

And as far as the lower piece of the sphere goes, I can't figure out what the limits on my triple integrals would be. I can find the volume by seperating it in to two pieces like this:

From the xy plane up to z=1/2: [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] r2 Sin[p] dr dp dt with theta from 0 to 2 pi, phi from 0 to pi/2, and row from 0 to 1/(2Cos[phi]).

From xy plane down (the lower hemisphere): [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] r2 Sin[p] dr dp dt with theta from 0 to 2 pi, phi from 0 to pi/2, and row from 0 to 1.

Add these together gives the volume, but what would I do to find the centroid of the entire piece?

By the way, r= row, p= phi, and t= theta
 

Answers and Replies

  • #2
HallsofIvy
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The unit sphere x2 + y2 + z2 =1 with density =1 is cut by the plane z=1/2. Find the volume and centroid of each piece.

As of now, i have (for the top piece) the integral of [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] r2 Sin[p] dr dp dt, with theta from 0 to 2 pi, phi from 0 to pi/2, and row from 1/(2Cos[phi]) to 1, but when i multiply in rCos[p] to find the z coordinate of the centroid, I get 3/8 which is below that part of the sphere.
[itex]\phi[/itex] does NOT go from 0 to [itex]\pi/2[/itex]. The plane z= 1/2 cuts the unit sphere where [itex]x^2+ y^2+ 1/4= 1[/itex] or [itex]x^2+ y^2= 3/4[/itex]. The distance from (0,0,0) to, say, [itex](\sqrt{3}{2}, 0, 1/2)[/itex] on that circle is 1 so we have [itex]\rho= 1[/itex] on that circle. Since [itex]x^2+ y^2= 3/4[/itex] in spherical coordinates is [itex]\rho^2 sin^2(\phi)= 3/4[/itex], with [itex]\rho= 1[/itex], we have [itex]sin^2(\phi)= 3/4[/itex] so [itex]sin(\phi)= \sqrt{3}/2[/itex]. [itex]\phi[/itex] goes from 0 to [itex]\pi/3[itex].

And as far as the lower piece of the sphere goes, I can't figure out what the limits on my triple integrals would be. I can find the volume by seperating it in to two pieces like this:

From the xy plane up to z=1/2: [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] r2 Sin[p] dr dp dt with theta from 0 to 2 pi, phi from 0 to pi/2, and row from 0 to 1/(2Cos[phi]).

From xy plane down (the lower hemisphere): [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] r2 Sin[p] dr dp dt with theta from 0 to 2 pi, phi from 0 to pi/2, and row from 0 to 1.

Add these together gives the volume, but what would I do to find the centroid of the entire piece?

By the way, r= row, p= phi, and t= theta
(The standard english transliteration of the Greek letter [itex]\rho[/itex] is "rho".)

Actually, a better way to do it is to integrate on the cone, [itex]\phi[/itex] from 0 to [itex]\pi/3[/itex], [itex]\rho[/itex] from 0 to [itex]1/(2cos(\phi))[/itex] and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex], then the rest of the sphere: [itex]\phi[/itex] from [itex]\pi/3[/itex] to [itex]\pi[/itex], [itex]\rho[/itex] from 0 to 1, and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex].

As for the centroid do exactly the same thing: multiply dV by [itex]z= \rho cos(\phi)[/itex], integrate over those two ranges and add. Then divide by the volume.
 

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