Recent content by richies

thank you so much, now I get it :D

So my interation will be like this, correct me if I am wrong r⊥ = h-z r sqr(z^2+(h-z)^2) M = integral from 0 to h of pi(h-z)^2*z^2dz * components: x=0 y=0 z = (1/M)*∫ (from 0->h) ∫(from0->2pi)∫(from 0->h)(sqr(z^2+(h-Z)^2))drdθdz.

"wouldn't the radius at height z be (h - z) ?" is that the radius at the base depends on the change in height ? and it is equal h - z ? if so, would the integral R = 1/M ∫p(r)rdV solve this problem?

Homework Statement Find the center of mass of the solid figure similar to a cone pointing upward with slope = 1 Note: the density varies with z^2 and the edge has a slope of 1. From symmetry we see that both Xc and Yc are equal to zero. Find the center of mass in the z direction as a function...