Question

Center of mass for the region bounded by the paraboloid z=4-x^2-y^2 and z=0 with p(x,y,z)=5-z

Answer 1

Well, youhave three integral sin your future. Can you find the first? Usually "M"?

Answer 2

first we have to find the mass. but i have trouble setting up the integrals

Answer 3

would it be \[\int\limits_{0}^{2\pi}\int\limits_{0}^{2}\int\limits_{0}^{4-r}(6-z) rdzdrd \Theta\]

Answer 4

Okay, first notice that z goes between:
\[0 \le z \le 4-(x^2+y^2)\]
To me this looks like a good time to switch to cylindrical. So we have:
\[0 \le z \le 4-r^2\]
\[0 \le \phi \le 2 \pi \]
Now we only need r.
So look at z=4-r^2 if z=0 r=2 and at the max height r will be zero. So we have what you have :D

Answer 5

except the top bound should be 4-r^2 not 4-r

Answer 6

thanks

Answer 7

malevolence19 is there easier way to find \[\frac{ }{ x }\]

Answer 8

instead of going through all of the integrals etc, since the difference is that you use only x

Answer 9

Well careful inspection tells you that x and y are both zero. By symmetry and the fact that:
\[x=r \cos(\phi);y=r \sin(\phi) \implies \iiint r \sin(\phi) \rho dV; \iiint r \cos(\phi) \rho dV\]
But if you realize that integrating sine or cosine over a full period (i.e., 2 pi) are just ZERO. That is:
\[\int\limits_0^{2 \pi}\sin(x)dx=\int\limits_0^{2 \pi} \cos(x)dx=0\]
So they are zero by symmetry and by the math.

Answer 10

Since z is just z is cylindrical you see that there is no sine or cosine so the integral is NON zero, as we'd expect.