# Problem determining p in triple integral

1. Mar 29, 2009

### gr3g1

The question states:

Find the center of mass of the solid that is bounded by the hemisphere z = sqrt(21 - x ^2 - y^2) and the plane z = 0 if the density at a point P is directly proportional to the distance from the xy-plane.

I know that the integral is setup :
$m = \int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{0}^{1} kzp^{2}sin\theta dpd\phid\theta$

How ever I do not see how p = 1 for this equation...

i know x^2 + y^2 + z^2 = p^2

any help would be appreciated... for all my other problems i would isolate x^2 + y^2 + z^2 and determine my p, however, for this problem, that isnt working...

thanks a lot
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 30, 2009

### HallsofIvy

Staff Emeritus
So "p" (actually the Greek letter "rho": $\rho$) is the radius of the sphere? If $z= \sqrt{21- x^2- y^2}$ then squareing both sides gives $z^2= 21- x^2- y^2$ or $x^2+ y^2+ z^2= 21$, a sphere of radius 21, not 1. Of course in the orginal form z cannot be negative so that is the upper hemisphere. If what you have is correct then the integral you want should be
$$\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2}\int_{\rho= 0}^{21} k\rho cos(\phi) \rho^2 sin(\phi) d\rho d\phi d\theta$$
Notice that I have made two corrections: in spherical coordinates, $z= \rho cos(\phi)$ and the differential involves $sin^2(\phi)$, not $sin^2(\theta)$. But it is true that the radius of the sphere, as given, is 21, not 1.

I wonder if there isn't a typo here and the formula was intended to be $z= \sqrt{1- x^2- y^2}$.