A thin sheet has the shape of the surface S. If its density (mass per unit area)
at the point $(x, y, z)$ is $ρ(x, y, z)$, then its center of mass is
$(x, y, z)$, where $x = 1/m$ (double integral) $(xρ(x, y, z) dS)$
$y = 1/m$ (double integral) $(yρ(x, y, z) dS)$
$z = 1/m$ (double integral) $(zρ(x, y, z) dS)$
and $m$ is the mass of the sheet. Find the center of mass of the sheet.
S is the part of the paraboloid $z = 12 − (1/6)x^2 − (1/6)y^2, z ≥ 0, ρ(x, y, z) = k$, where $k$ is a constant.
Any help on this would be appreciated.
$x = r\cos t\\ y =r \sin t\\ z = 12 - \frac 16 r^2$
$dS = \|(\frac {\partial x}{\partial r},\frac {\partial y}{\partial r},\frac {\partial z}{\partial r}) \times(\frac {\partial x}{\partial t},\frac {\partial y}{\partial t},\frac {\partial z}{\partial t}) \| = r\sqrt{\frac {r^2}{9} +1}\ dr\ dt$
$m = \int_0^{2\pi} \int_0^\sqrt{72} \rho(r,t) r \sqrt {\frac {r^2}{9} + 1} \ dr \ dt\\ 2\pi k \int_0^\sqrt{72} r \sqrt {\frac {r^2}{9} + 1} \ dr\\ u = \frac {r^2}{9} + 1\\ du = \frac 29 r\\ (2\pi k) \int_1^9 \frac {9}{2}\sqrt u \ du\\ (2\pi k) 3u^{\frac {3}{2}}|_1^9\\ (2\pi k) 3(26)$
$\bar z = \frac {2\pi k}{m} \int_0^\sqrt{72} r (12-\frac 16 r^2) \sqrt {\frac {r^2}{9} + 1}\ dr$
Do the indentical u-sub as above.
$\bar z = \frac {1}{78}\int_1^9 \frac {27}{4}(9-u)\sqrt u \ du$
$\bar x = $$\int_0^{2\pi} \int_0^\sqrt{72} r^2\cos t \sqrt {\frac {r^2}{9} + 1} \ dr \ dt\\ \int_0^{2\pi} \cos t\ dt \int_0^\sqrt{72} r^2\sqrt {\frac {r^2}{9} + 1} \ dr = 0$
$\bar y = 0$