# Thread: help starting a triple integral/center of mass problem

1. ## help starting a triple integral/center of mass problem

i need help setting up the integral expressions to compute the centroid please!
the solid bounded by the paraboloids z=1+x^2+y^2 and z=5-x^2-y^2 with density proportional to the distance from the plane z=5
so i need to find the boundaries for x,y, and z, which i think i can do if i set x=0 then y=0?
my main problem is figuring out how to put the density into an equation. i think it has to do with the distance formula. f(x,y,z)= k* sqrt of 5? any hints would help a ton! thanks!

2. I tried to do this problem last night but ended up getting swamped by calculations. One way to find the answer is to switch to cylindrical coordinates:

\displaystyle \begin{aligned} x&=r\cos\theta\\ y&=r\sin\theta\\ z&=z \end{aligned}

so that the equations for the paraboloids become

$\displaystyle z=1+r^2$
$\displaystyle z=5-r^2.$

Solving for the intersection of the two gives us

\displaystyle \begin{aligned} 1+r^2&=5-r^2\\ 2r^2&=4\\ r^2&=2\\ r&=\sqrt{2} \end{aligned}

for the radius of a circle in the $\displaystyle xy$-plane that bounds the solid. Our center of mass for the $\displaystyle z$-coordinate becomes

$\displaystyle z_M=\frac{1}{V}\int_0^{2\pi}\int_0^{\sqrt{2}}\int_ {1+r^2}^{5-r^2} z\cdot k(5-z)\cdot r\,dz\,dr\,d\theta,$

where $\displaystyle V$ is the volume of the solid.

Hint: The $\displaystyle x$- and $\displaystyle y$-coordinates of the center of mass must be zero. Why is this?