Center of mass problem

dawgs1234

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Dec 16, 2010
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I'm having problems solving this. Any help would be appreciated! I know that xbar and ybar are 0

Find the center of mass of the solid inside the sphere ? = 2, below the cone
? = ?/3 and above the plane z = 0, if the density is proportional to the distance from
the origin. Use symmetry where possible.
 
Did you manage to get your triple integral set up?.

02π0π302ρ2sin(ϕ)dρdϕdθ\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{2}{\rho}^{2}sin({\phi})d{\rho}d{\phi}d{\theta}

The mass is M=kρdV\displaystyle M=\int\int\int k{\rho}dV

As you stated, because of symmetry x=y=0\displaystyle \overline{x}=\overline{y}=0.

So, Mz=0=k02π0π302ρcos(ϕ)zρρ2sin(ϕ)dρdϕdθ\displaystyle M_{z=0}=k\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{2}\underbrace{{\rho}cos(\phi)}_{\text{z}}\cdot {\rho}\cdot {\rho}^{2}sin(\phi)d{\rho}d{\phi}d{\theta}
 
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