How Do You Calculate the Volume and Centroid of a Sphere Cut by a Plane?

[simpleman008]

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 \int\int\int 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 into two pieces like this:

From the xy plane up to z=1/2: \int\int\int 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): \int\int\int 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

 

[HallsofIvy]

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 \int\int\int 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.

\phi does NOT go from 0 to \pi/2. The plane z= 1/2 cuts the unit sphere where x^2+ y^2+ 1/4= 1 or x^2+ y^2= 3/4. The distance from (0,0,0) to, say, (\sqrt{3}{2}, 0, 1/2) on that circle is 1 so we have \rho= 1 on that circle. Since x^2+ y^2= 3/4 in spherical coordinates is \rho^2 sin^2(\phi)= 3/4, with \rho= 1, we have sin^2(\phi)= 3/4 so sin(\phi)= \sqrt{3}/2. \phi goes from 0 to \pi/3.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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 into two pieces like this:<br /> <br /> From the xy plane up to z=1/2: \int\int\int r<sup>2</sup> 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]).<br /> <br /> From xy plane down (the lower hemisphere): \int\int\int r<sup>2</sup> Sin[p] dr dp dt with theta from 0 to 2 pi, phi from 0 to pi/2, and row from 0 to 1.<br /> <br /> Add these together gives the volume, but what would I do to find the centroid of the entire piece?<br /> <br /> By the way, r= row, p= phi, and t= theta </div> </div> </blockquote>(The standard english transliteration of the Greek letter \rho is "rho".)<br /> <br /> Actually, a better way to do it is to integrate on the <b>cone</b>, \phi from 0 to \pi/3, \rho from 0 to 1/(2cos(\phi)) and \theta from 0 to 2\pi, then the rest of the sphere: \phi from \pi/3 to \pi, \rho from 0 to 1, and \theta from 0 to 2\pi. <br /> <br /> As for the centroid do exactly the same thing: multiply dV by z= \rho cos(\phi), integrate over those two ranges and add. Then divide by the volume.