# Mass and center of mass

### Mass and center of mass

Find the mass and center of mass of the object corresponding to the given surface and mass density. The portion of the plane $$3x + 2y + z = 6$$ inside the cylinder $$x^2 + y^2 = 4, \rho (x, y, z) = x^2 + 1.$$

Any help would be appreciated!
mario99

Posts: 3
Joined: Wed Aug 19, 2020 4:45 am
Reputation: 0

### Re: mass and center of mass

$$m = \iint\limits_S \rho \ dS = \iint\limits_R (x^2 + 1) \sqrt{\left(\frac{\partial z}{\partial x} \right)^2 + \left(\frac{\partial z}{\partial y} \right)^2 + 1} \ \ dx \ dy = \iint\limits_R (x^2 + 1) \sqrt{\left(-3 \right)^2 + \left(-2 \right)^2 + 1} \ \ dx \ dy = \sqrt{14}\iint\limits_R (x^2 + 1) \ \ dx \ dy = \sqrt{14}\int_{0}^{2\pi} \int_{0}^{2} ((r\cos \theta)^2 + 1) \ r \ dr \ d\theta = \sqrt{14}\int_{0}^{2\pi} \int_{0}^{2} (r^3\cos^{2} \theta + r) \ dr \ d\theta = \sqrt{14}\int_{0}^{2\pi} (4\cos^{2} \theta + 2) \ d\theta = 8\pi \sqrt{14}$$

by symmetry $$\overline{x}$$ = $$\overline{y} = 0$$

$$\overline{z} = \frac{1}{m}\iint\limits_S z\rho \ dS = \frac{\sqrt{14}}{8\pi\sqrt{14}}\iint\limits_R (6 - 3x - 2y)(x^2 + 1) \ dx \ dy = \frac{1}{8\pi}\int_{0}^{2\pi} \int_{0}^{2} (6 - 3r\cos \theta - 2r\sin \theta)((r\cos \theta)^2 + 1) \ r \ dr \ d\theta = \frac{1}{8\pi}\int_{0}^{2\pi} \int_{0}^{2} (6 - 3r\cos \theta - 2r\sin \theta)(r^3 cos^2 \theta + r) \ dr \ d\theta = \frac{1}{8\pi}\int_{0}^{2\pi} \int_{0}^{2} (6r^3 \cos^{2} \theta - 3r^4\cos^{3} \theta - 2r^4\sin \theta \cos^{2} \theta + 6r - 3r^2 \cos \theta - 2r^2 \sin \theta) \ dr \ d\theta = \frac{1}{8\pi}\int_{0}^{2\pi} (24 \cos^{2} \theta - \frac{96}{5}\cos^{3} \theta - \frac{64}{5}\sin \theta \cos^{2} \theta + 12 - 8 \cos \theta - \frac{16}{3} \sin \theta) \ d\theta = \frac{1}{8\pi} \cdot 48\pi = 6$$
Guest

Done by

-Jambo
Guest