Charge density in concentric spheres Question: 

If there are two conducting spherical shells and the inner shell is grounded, what will be >the charge density in the inner shell if there is a charge Q placed on the outer shell?

Yes, this is a HW problem, but I am not asking you guys to solve it for me... just show me the way :D
If there is a charge on the outer shell with radius $a$, the charge density will be $\frac{Q}{4\pi a^2}$.
That should induce, a charge density on the inner side of outer shell -- $\frac{Q}{4\pi (a-x)^2}$  where $x$ is thickness of the outer shell. 
Now, if the inner shell was never grounded, $\frac{Q}{4\pi b^2}$ ($b$ = radius of the inner shell) charge density would have been induced on the inner shell, now that it is grounded there would be no charge on it, right?
There is no need for a charge to be induced for the electric field to be 0 inside the inner shell.
This seems to be my conclusion, but I fear it is too simple for the question. There might be something I am missing. 
Edit: the hint of the problem says when a sphere is grounded, potential is infinity.
 A: 
Now, if the inner shell was never grounded, $\frac{Q}{4\pi b^2}$ ($b$ = radius of the inner shell) charge density would have been induced on the inner shell

If the inner shell were not grounded, it wouldn't be connected to anything. And if it wasn't connected to anything, how would it be able to have any charge density other than zero? Where would the extra charge come from?

now that it is grounded there would be no charge on it, right?

No, as pointed out in the comments, the potential is zero, but that doesn't mean the charge is zero.
Can you figure out the potential of the outer sphere? Once you know the potentials of both spheres, what else can you calculate using that information?
A: As the problem is spherically symmetric, you know that the potential must go as $1/r$ in free space. Let $Q$ be the charge at the outer sphere and $Q'$ the charge at the inner sphere; then the potential must be expressed as $\frac{A\,(Q+Q')}{r}$ for the space outside the outer shell ($r>a$) and as $\frac{A\,Q'}{r}+K$ at the space between the inner shell and the outer shell ($a>r>b$).
You can get the value of $Q'$ and $K$ by matching both expressions for the potential at $r=a$ (the potential doesn't jump at surface charge distributions) and by forcing a zero potential at $r=b$ (you have two equations with two unknowns).
A: Firstly assume that you know the charge on the inner shell $q_{in}$. Then calculate the electric field between the shells $E_{in}$ and the field outside shells $E_{out}$ by using Gauss law. Know by knowing that the potential at the $\infty$ is 0 and that when conductor is grounded has the same potential, you can write potential difference $U$ in two ways:
$$U = \int_R^\infty E_{out}\,dr$$
$$U = -\int_r^RE_{in} \, dr$$
Where $r$ is inner but R is outer shell radius.
