# Thread: triple integral and center of mass

1. ## triple integral and center of mass

hi there,

my question is:

i have a D = {(x,z): ax^2 <= z <= h)}, where a and h are parameters.

if you take this D and rotate it around the z axis,

you get a shape in r^3 - which we'll call V.

we start changing the a parameter, so V changes.

how do i need to change h so that the z coordinate of the center of

mass remains in the same point (the body's mass density is constant)?

i realized that the shape of V is a paraboloid whose outer "radius" is sqrt(h/a) and runs on z from 0 to h. i've tried to calculate a
triple integral of z*dx*dy*dz / the triple integral of 1*dx*dy*dz, but get the answer wrong.

i know my description of the question is somewhat long,

2. The lower boundary of the volume of revolution $V$ of $z=ax^2$ about the $z$-axis is given by

$z=ar^2\quad\quad 0\le z \le h.$

You are correct that to find the $z$-coordinate of the center of mass of $V$, $z_M$, we divide

$\int\int\int z\,dx\,dy\,dz\quad\quad\mbox{by}\quad\quad\int\int \int\,dx\,dy\,dz.$

To find each, we may convert to cylindrical coordinates:

$\int_0^{2\pi}\int_0^{\sqrt{\frac{h}{a}}}\int_{ar^2 }^h z\,dz\,dr\,d\theta,$

or, alternatively, use the Disc Method:

$\int_0^h \pi\left(\sqrt{\frac{z}{a}}\right)^2z\,dz.$

I found the same answer for each, but it did not depend on the value of $a$.