Help finding mass and center of mass?

What am I doing wrong in this problem involving finding the center of mass of a uniform hollow hemispherical bowl?

• I am considering a differential ring at a height of z from the center, then calculating the differential mass dm and then using the definition of center of mass I am integrating zdm from 0 to R(radius of bowl)and then doing the requisite calculations. In this way the answer is turning out to be R/3, which is wrong and should be R/2. Where am I doing wrong? Total area A=2piR^2, total mass=M Mass per unit area m=M/A=M/(2piR^2) For the differential ring, dA=2pixdx (x is the radius of differential ring) dm=mdA=m(2pix)dx Now from geometry we have z^2=R^2-x^2 Hence the center of mass along z axis is given by (Integration from 0 to R (z dm))/M. Solving and simplifying this yields R/3.

• Answer:

  The setting up of the integral and the calculation part are okay. The problem is in modeling. According to your approach, you have sliced the sphere parallel to its base. Now approximating each of the slices as a cylinder is not correct - as you go away from the center (z increases), the approximation keeps worsening. A slice is more like a frustum, and not a cylinder. Before going on to the correct procedure, let me give a little more insight into the error: Close to z=R, the slice of the hemisphere is almost like a disc, while you have approximated it as a cylinder. So the approximation of the area is lesser than actual. This happens for all slices close to z=R. Hence, they contribute less in the integral than they actually should have, and consequently you get the center of mass closer to the center. One correct way is to use spherical coordinates to find the area of the frustum: (Rectangular coordinates make the calculations a lot messier.) Let the slice be between [math]\theta[/math] and [math]\theta + d\theta[/math] angle from the base. Then the slant height of the frustum is [math]l = R d\theta[/math] The radii are [math]r1 = R cos\theta[/math] [math]r2 = R cos(\theta+d\theta)[/math] So the surface area is [math]dS = \pi (r1 + r2) L[/math] Evaluating the integral using this gives the correct answer.