Shell method and volumes Find the volume of a cap in a sphere of radius $r$, with a height $h$

I have solved the problem using the washer method (or the disk method), but I'm currently stuck as when I'm trying to solve it using cylindrical shells. I have that (let c be the circumference of a typical cylinder):
$c = 2\pi x$
$h = (r^{2} - x^{2})^{1/2} - r + h$
And the bounds of integration would be from $x = 0$ to $x = (2hr - h^{2})^{1/2}$.
Or maybe I'm not visualizing the problem correctly. I was as well wondering, when dealing with the washer method for finding the volumes, we look at the inner and outer radius from the cross section of the solid (perpendicular to the axis we are rotating it) or from the region R we are rotating it? 
Thank you for everything.
Note: A picture to make my words clearer: http://en.wikipedia.org/wiki/File:Spherical_Cap.svg
 A: the volume of the cap using cylindrical shells is 
$$V = 2\pi \int_0^{2rh - h^2} x(y - r + h) \ dx $$ subject to the constraint $x^2 + y^2 = r^2, x \ge 0, y \ge 0.$ Use change of variable $x = r \sin t, y = r \cos t$ to evaluate the integral.  
$\begin{eqnarray} 
V &=& 2 \pi \int_0^\alpha r\sin t(r \cos t - r + h) r \cos t \ dt, \cos(\alpha) = {r-h \over r}\\
&=&2\pi \int_0^\alpha[r^3\cos^2t \sin t -(r-h)\cos t \sin t]\ dt \\
&=&2\pi[-{1 \over 3}r^3\cos^3 t+{1 \over 2}r^2(r-h)\cos^2 t]_0^\alpha\\
&=&2\pi[{1 \over 3}r^3 - {1 \over 2}r^2(r-h) -{1 \over 3} (r-h)^3+{1 \over 2}(r-h)^3\\
&=&{1 \over 3} \pi[2r^3 -(r-h)(2r^2 + 2rh - h^2)]
\end{eqnarray}$
the formula checks for $h = 0, V = 0$ and $h = r, V = {2 \over 3}\pi r^3.$
A: Ok, first we got a sphere with outer Radius $r_o$ and inner radius $r_i$. Also we just want the volume of a cap with given height $h$. Let's assume this height is measured from the northpole as in the picture.
We know the radius $R(x)$ of the circle at a given height $x$ is:
$$ R(x) = \sqrt{r^2-x^2} $$
So the formulas for the outer radius $R_o(x)$ and the inner radius $R_i(x)$ are:
$$ R_o(x) = \sqrt{r_o^2-x^2} \quad \quad R_i(x)=\sqrt{r_i^2-x^2} $$
Now we have a problem, if $r_o > x > r_i$ the inner circle will have no effect! as there is no inner circle anymore.
So we have to split the integral into two. Let $a$ be the height where the inner circle does not exists than it is given throught $a=r_o-r_i$ giving you the integrals:
$$ V = \pi\left(\ \int\limits_a^r R_o^2(x)\ \text{d}x+\int\limits_h^{r-a}R_o^2(x) -R_i^2(x)\ \text{d} x\ \right )$$
If you look at the second integral you will notice that you are just calculating the volume of the inner (empty) sphere and substract it from the outer sphere!
