# Grav field fun!

by Phymath
Tags: field, grav
 P: 186 $$\vec{g}(x,y,z) = -kG((x^3 y^2 z^2)\hat{e_x} + (x^2 y^3 z^2)\hat{e_y} + (x^2 y^2 z^3) \hat{e_z})$$ given this grav field (k is constant) find the mass density of the source of this field, and what is the total mass in a cube of side 2a centered about the origin? hmmm well we all know...$$\int \int \vec{g} \bullet d\vec{a} = 4 \pi G m_{enclosed}$$ and density $$p = \frac{m}{V}$$ at least the overall density of it is (non-differential) sooooo...$$\frac{1}{4 \pi G}\int\int\int (\nabla \bullet \vec{g}) dV = m_{source/enclosed}$$ now the limits i made a cube of side 2a, because the flux through a box is easier when g is given in cart coords....any way i get... $$m_{source} = \frac{-2}{3 \pi}G k a^9$$ how do I get a neg mass (unless this is dark matter which it very well could be) and I'm thinking i missed something about density cause why would it ask that first and then the mass enclosed second?...a lil help?
 Sci Advisor HW Helper P: 11,928 Grav field fun! You forgot about the minus in the RHS of Gauss's theorem for the gravitostatic field... $$\oint\oint_{\Sigma} \vec{g}\cdot d\vec{S}=\mbox{-}4\pi G m_{\mbox{enclosed by}\ \Sigma}$$ Daniel.
 P: 186 ok if $$\frac{1}{4 \pi G}\int\int\int (\nabla \bullet \vec{g}) dV = m_{source/enclosed}$$ then...ill just do it and show you $$\nabla \bullet \vec{g} = -9kGx^2y^2z^2$$ $$\frac{9k}{4 \pi }\int^a_{-a}\int^a_{-a}\int^a_{-a} (x^2y^2z^2) dxdydz = \frac{2}{3}k \pi a^9 = m_{source/enclosed}$$ so as my "flux cube" getts larger obviously it isn't limiting at some value as the function is ever increasing so, the mass should always be increasing yes?