Conserved quantity from a conserved current

In my QFT course, it was asserted that the conserved quantity associated with some conserved current is given by $$Q = \int_v j^0d^3x$$ where $$j^0$$ is the time component of the conserved current, and $$d^3x$$ is over the three spatial coordinates. My professor "left it as an exercise to the reader" to come up with an explanation for this, but I'm having trouble figuring out where this comes from.

My very hacky way of thinking about it currently is if you say $$j^\mu = (j^0,J)$$, where $$j^0$$ is the time component and $$J$$ comprises the spatial components. If you expand out $$\partial_\mu j^\mu = 0$$, you get $$\partial_tj^0 -\nabla\cdot J = 0$$. And if you integrate with respect to both time and space, you get

$$\int d^3x j^0 = \int d^3x dt( \nabla\cdot J)$$

Where the RHS is the conserved quantity, and the LHS is the form we want it in. I understand the RHS has form where you can use the divergence theorem, but I wouldn't then be sure where to take it. And I'm not even sure if this approach is right. Why would we say the RHS is a conserved quantity?

Can someone provide a bit of help on this one? Sorry if my notation isn't standard, I am still learning the lingo.

Update

Ok, so I see how one might be able to interpret the RHS, $$\int d^3x dt( \nabla\cdot J)$$ as a conserved quantity. By definition, if you take the time derivative, it must equal zero. So, let's see.

$$\frac{d}{dt}\int d^3x dt( \nabla\cdot J) = \int d^3x( \nabla\cdot J)$$

and then using the divergence theorem

$$\int d^3x( \nabla\cdot J) = \int_S(J\cdot \vec{n})dS = 0?$$

Should the spatial components of the conserved current vanish at the boundary? If not, I am unsure why this should be zero.

We mame use of $$\partial_\mu \mathcal j^\mu=0$$ $$\int d^3x (\partial_0\mathcal j^0 +\nabla\cdot \bf J)=0$$ for big enough volume the divergence term vanishes (by the divergence theorom $$\int d^3x \nabla \cdot \bf J=\int \bf J \cdot \bf \hat n dA$$ and the boundaries at infinity vanishes) giving you $$\frac{d}{dt}\int d^3x \mathcal j^0=0$$ Note that we make the boundaries at infinity vanish all the time. Assumption of vanishing boundaries at large enough boundaries generally don't matter as we observe the vicinity of fields is quite small. The irony is we ignore them in General Relativity too, even though vicinity of gravitational fields are quite large but at a far enough distance flat spacetime will be observed.