Triple integrals with polar coordinates. I have a following integral: $$\int_0^1 dx\int_0^{\sqrt{1-x^2}}dy \int_\sqrt{x^2+y^2}^\sqrt{1-x^2-y^2}z^2dz$$
Which i have to solve by introducing polar coordinates, which is, by itself, relatively simple:
$$x=\rho\cos\theta\sin\phi \\ y=\rho\sin\theta\sin\phi \\ z=\rho\cos\phi$$
Besides this, i need to find Jacobian since i introduced a substitution, and since this is well known substitution Jacobian is $$J=\rho^2\sin\phi$$
Now, since i introduced polar coordinates, bounds of integral should be in polar form too, lower bound of the first, $dz$ integral, is simple $$\sqrt{x^2+y^2}= \rho\sin\phi$$, but i don't know what to do with this expression $$\sqrt{1-x^2}=\sqrt{1-\rho^2\cos^2\theta\sin^2\phi}$$
since, after introducing polar coordinates, this bound has all of the variables in itself, which makes it impossible to integrate over any of the variables i have, so i don't know how to solve this. Any help appreciated.

Actually, this is a volume integral of the form $\iiint_V f(x,y,z) dx dy dz$ ; where V is the volume(can be define by the limits of the given  triple integration).
Since you want to solve this by using polar co-ordinate system ,so you need to know the limits of $\rho$ ,$\theta$ and $\phi$. For this you have to understand about $\rho$=C(sphere of radius C and centre at origin) , $\phi$=C(pair of cone with semi-vertical angle C and $z-axis$ as its axis) and $\theta$=C(pair of straight lines passing through origin) ,C is any constant.
Here, $V$ is the volume in the $+ve$ part of $ z-axis$(only in the first octant) , enclosed between $\rho=1$ and $\phi$=$\pi/4$.
Therefore, the limits are-
$\rho$: $0\to1$
$\phi$: $0\to\pi/4$
$\theta$:$  0\to \pi/2$
so, the integral can be given as follows,
$=$$\int \rho^4 d\rho$ $\int\sin\phi \cos^2\phi$ $d\phi$ $\int d\theta$.
After applying limits , you'll get the answer. 
