Question

Find the mass and center of gravity of the cylindrical solid that has density δ(x, y, z) = h − z and is enclosed by

x

2 + y

2 = a

2

, z = 0, and z = h

Answer 1

$mass,m=\int\int\int\delta(x,y,z)dV\\ =\int_{x=-a}^a\int_{y=-a}^a\int_0^h(h-z)dzdydx\\ =\int_{x=-a}^a\int_{y=-a}^a[hz-\frac{z^2}{2}]_0^hdydx\\ =\frac{h^2}{2}\int_{x=-a}^a\int_{x=-a}^a dydx\\ =\frac{h^2}{2}\int_{x=-a}^a[y]_{-a}^a dx\\ =a{h^2}\int_{x=-a}^a dx\\ =a{h^2}[x]_{-a}^a\\ =2a^2h^2\\ \text{Now, calculate moments}\newline M_{xy}=\int\int \int z\delta dV =\int_{x=-a}^a\int_{y=-a}^a\int_0^h hz-z^2dzdxdy\\ =\int_{x=-a}^a\int_{y=-a}^a[\frac{hz^2}{2}-\frac{z^3}{3}]_0^hdydx\\ =\frac{h^3}{6}\int_{x=-a}^a\int_{x=-a}^a dydx\\ =\int_{x=-a}^a[y]_{-a}^a dx\\ =a\frac{h^3}{6}\int_{x=-a}^a dx\\ =a\frac{h^3}{6}[x]_{-a}^a\\ =2a^2\frac{h^3}{6} \\ M_{yz}=\int\int \int x\delta dV =\int_{x=-a}^a\int_{y=-a}^a\int_0^h x(h-z)dzdxdy\\ =\int_{x=-a}^a\int_{y=-a}^ax[hz-\frac{z^2}{2}]_0^hdydx\\ =\frac{h^2}{2}\int_{x=-a}^a\int_{x=-a}^a xdydx\\ =\frac{h^2}{2}\int_{x=-a}^ax[y]_{-a}^a dx\\ =a{h^2}\int_{x=-a}^a xdx\\ =a{h^2}[\frac{x^2}{2}]_{-a}^a\\ =0\\ M_{xz}=\int\int \int y\delta dV =\int_{x=-a}^a\int_{y=-a}^a\int_0^h y(h-z)dzdxdy\\ =\int_{x=-a}^a\int_{y=-a}^ay[hz-\frac{z^2}{2}]_0^hdydx\\ =\frac{h^2}{2}\int_{x=-a}^a\int_{x=-a}^a ydydx\\ =\frac{h^2}{2}\int_{x=-a}^a[\frac{y^2}{2}]_{-a}^a dx\\ =a{h^2}\int_{x=-a}^a 0dx\\ =0\\ \text{Center of gravity is given by}(x,y,z)then,\\ x=\frac{M_{yz}}{m}=\frac{0}{2a^2h^2}=0,\\ y=\frac{M_{xz}}{m}=\frac{0}{2a^2h^2}=0,\\ z=\frac{M_{xy}}{m}=\frac{2a^2\frac{h^3}{6} }{2a^2h^2}=\frac{h}{6} \\$