# Finding center of mass of solid

• Draconifors

## Homework Statement

A solid B occupies the region of space above ##z=0## and between the spheres ##x^2 + y^2 + z^2 = 16## and ##x^2+y^2+(z-1^2) = 1##. The density of B is equal to the distance from its base, which is ##z = 0##. The mass of the solid B is ##\frac{188\pi}{3}##. Find the coordinates of the center of mass.

## Homework Equations

Center of mass in z = ##\frac{1}{m} \iiint \limits_E z *\rho(x,y,z) dV##

## The Attempt at a Solution

The center of mass in x and y is at coordinate 0 because of the symmetry of the domain, so there's only the z-value to calculate.

So my integral is ## \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{2\cos\phi}^{4} \rho \cos\phi * \rho \cos\phi * \rho^2\sin\phi d\rho d\phi d\theta##, then divided by the mass, but the answer sheet (which is a student's graded paper) has ##\rho\cos\phi * \rho^2\sin\phi## as the term to integrate, and therein lies my question: why are they only integrating z once? Don't we have z (the density) * z (the term already present in the center of mass equation) as the term to integrate?

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## Homework Statement

A solid B occupies the region of space above ##z=0## and between the spheres ##x^2 + y^2 + z^2 = 16## and ##x^2+y^2+(z-1^2) = 1##. The density of B is equal to the distance from its base, which is ##z = 0##. The mass of the solid B is ##\frac{188\pi}{3}##. Find the coordinates of the center of mass.

## Homework Equations

Center of mass in z = ##\frac{1}{m} \iiint \limits_E z *\ rho(x,y,z) dV##

## The Attempt at a Solution

The center of mass in x and y is at coordinate 0 because of the symmetry of the domain, so there's only the z-value to calculate.

So my integral is ## \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{2\cos\phi}^{4} \rho \cos\phi * \rho \cos\phi * \rho^2\sin\phi d\rho d\phi d\theta##, then divided by the mass, but the answer sheet (which is a student's graded paper) has ##\rho\cos\phi * \rho^2\sin\phi## as the term to integrate, and therein lies my question: why are they only integrating z once? Don't we have z (the density) * z (the term already present in the center of mass equation) as the term to integrate?

You are doing it the hard way, given the symmetry you already noticed. Between ##z## and ##z + \Delta z## you have a thin circular disk centered at ##(0,0,z)## whose mass you can get because you can figure out the radius of the disc. (There will be two separate formulas for the radius, depending on whether the disc extends out to the lower surface or the upper surface). If ##m(z) \, dz## is the mass of the thin circular disc of thickness ##dz## at ##z##, the CM is just ##\frac{1}{M} \int_0^4 z\, m(z) \, dz,## where ##M## is the mass of the whole solid.