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Sorry for boring you my friends. I wonder what a bounded closed volume in $\mathbb{R}^3$ looks like, if the volumic integral on this volume meets the following condition: $$ \int_Vx\text{dV}= \int_Vy\text{dV}= \int_Vz\text{dV} = 0 $$ In my opinion, an axi-symmetric volume placed at the origin of $\mathbb{R}^3$ coinciding in its gravity center with its axis of symmetry parapllel to the one of three axes of $\mathbb{R}^3$. It is obvious not the only case. I would like to know if there is other type of volume and what they look like. Thank you in advance for taking a glance and giving some hint.

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When these integrals do vanish, it is clear that the origin is a center of gravity of the body $V$, and vice versa. It seems to me this is the most general condition on $V$ you are looking for.

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    Firstly, thank you for your answer. I still has a little question: when $\int_Vx\text{dV}= \int_Vy\text{dV}= \int_Vz\text{dV} = 0$ meets the definition of center of mass $\int_V\left(x+y+z\right)\text{dV}$ which indicates that the gravity center locates at the origin, it's OK. However, if a volume meets gravity center condition, it may be violate the three individual conditions. What's your opinion on this? – 2017-01-11
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    The coordinates of the gravity center (also mass center in this case) are $$x=\frac{1}{\text{vol}(V)}\int_V x\text{d} V,\;y=\frac{1}{\text{vol}(V)}\int_V y\text{d} V,\;z=\frac{1}{\text{vol}(V)}\int_V z\text{d} V$$ That is why vanishing of all these integrals is equivalent to the loaction of the center of mass at the origin. – 2017-01-11