Volume of intersection between a cone and cylinder Find the volume above the x-y plane inside the cone $z=2-(x^{2}+y^{2})^{1/2}$ and inside the cylinder $(x-1)^{2} + y^{2}=1$

Now using calculus this is actually a rather difficult integration using a Df matrix the bounds are rather un-intuitive to find, a simple substitution to polar actually doesn't make ur any life easier. ( you end up have to integrate $2-(1+r\cos(\theta)+r^{2})^{1/2}rdrd\theta$ in square cords u have $\int^{1}  \int^{(1-(x-1)^{2})^{1/2}} (x^{2}+y^{2})^{1/2}dydx$ and although there is a formula in most calc text books to integrate this ingratiating again in x will be very ugly after the sub ( the upper bound isn't actually right but it has something like that in it each integral is evaluated 0 to upper bound)
$x-1=r\cos(\theta)$ and $y=r\sin(\theta)$ was the sub i used
curiously can anyone solve this without calculating a bound analytically merely by integrating up that surface in 1 step? ( no cheating and breaking it apart into 2 integrals ie this minus this or this plus this doubles are fine and i am fine with an integration by parts trick.) other then that any mathematical trickery is welcome.
Edit
Understand the answer.
 A: Using Maple with $$\int _{0}^{2}\!\int _{-\sqrt {1- \left( x-1 \right) ^{2}}}^{\sqrt {1-
 \left( x-1 \right) ^{2}}}\! \left(2-\sqrt {{x}^{2}+{y}^{2}}\right) {dy}\,{dx}
$$
it looks like the answer is $2 \pi - 32/9$, certainly not $2 \pi/3$.
EDIT: Actually, the easy way to do this is to use polar coordinates, noting that the equation of the cylinder is $r = 2 \cos(\theta)$ for $-\pi/2 < \theta < \pi/2$.  The
integral becomes 
$$ \int_{-\pi/2}^{\pi/2} \int_{0}^{2 \cos(\theta)} r (2 - r)\ dr \ d\theta$$
A: Parametrization of the cylinder is 
$$
\begin{align*}
x-1 &= \cos \theta \\
y &= \sin \theta \\
z &= r
\end{align*}
$$
To find the curve that's intersection of the cylinder and the cone, just substitute equations above to the equation of cone
$$
z = 2-\sqrt{x^2+y^2} = 2-\sqrt{\left( \cos\theta+1\right)^2+\sin^2\theta} = 2-\sqrt{\cos^2\theta+\sin^2\theta+2\cos\theta+1} = \\
=2-\sqrt{2+2\cos\theta} = 2-\sqrt{4\cos^2\frac \theta 2}  = 2\left(1\pm\cos\frac\theta 2\right)
$$
So, polar coordinates of that curve is
$$
\begin{align*}
x &= \cos\theta+1\\
y &= \sin\theta \\
z &= 2\left(1\pm\cos\frac\theta 2\right)
\end{align*}
$$
It can be easily checked by Mathematica's ParametricPlot3D function 

So now, to find volume of this body, you need to find appropriate integration limits for the triple integral
$$
V = \int_0^{z_0} \int_0^{\theta_0} \int_0^{r_0} rdr\,d\theta\,dz
$$
where $z_0, \theta_0(z), r_0(z,\theta)$ are some functions, that need to be found based on the geometry of the body.
PS: I, personally, didn't find those limits yet, and this post is more of a hint, than a solution.
