Tough integral related to the volume of a set Let $A = B(0,1) \cap \{ (x,y,z) : y^2 + z^2 \leq \frac{1}{4} \} $, where $B(0,1) = \{ (x,y,z) : x^2+y^2+z^2 < 1 \} $
How can I compute the volume of $A$?? My idea is to use triple integrals, but I have no idea how to set up such an integral. I hope to obtain some help. thanks
 A: This volume is evaluable using various methods: an elementary integration or a "disk integration" technique, setting up a triple integral. 
The volume of the solid region $A$ generated by the intersection between the sphere and cylinder is found adding up the volumes of the cylinder of radius $R = 1/2$ and height $\sqrt{3}$ and the two identical "caps", the one situated in the $x$  positive half-space (let's indicate it with $T$) being described by the following relations:
$$
T = \left\{(x,y,z)\in\mathbb{R}^3\;\bigg|\;x^2+y^2+z^2\leq 1,\;
\; 
\frac{\sqrt{3}}{2}\leq x\leq 1\right\}
$$
that is we have to evaluate the following sum:
$$
V=\pi\cdot\left(\frac{1}{2}\right)^2\cdot\sqrt{3}
+
2\cdot\iiint_T\text{d}x\,\text{d}y\,\text{d}z\;.
$$
Using an elementary integral instead of a triple integral, the problem of finding the volume of one single "cap" requires to evaluate the volume of the "solid body" obtained rotating the region 
$$
T'=\left\{
(x,y)\in\mathbb{R}^2\,\left|\right.\;\frac{\sqrt{3}}{2}\leq x\leq 1,\;
0\leq y\leq\sqrt{1-x^2}
\right\}
$$
around the $x$-axis, and keeping in mind the relation $\int_a^b\pi\big[f(x)\big]^2\,\text{d}x$ for the volume of a solid of rotation, you can easily write
$$
V=\frac{\pi\sqrt{3}}{4}+2\int_{\sqrt{3}/2}^{1}\pi\cdot(1-x^2)\,\text{d}x\;.
$$ 
On the other hand, if we use a triple integration, we can imagine to "cut" the spere with planes $x=x_0$ parallel to the $yOz$ plane, noting that all of these "slices" are circles of radius $\sqrt{1-x_0^2}\,$, having area
$$
S(x_0)=\pi\cdot(1-x_0^2)\,,\qquad\forall x_0\in\left[\frac{\sqrt{3}}{2},1\right].
$$
Integrating over the interval $\frac{\sqrt{3}}{2}\leq x\leq 1 $ we can simply pose
$$
2\cdot\iiint_T\;\,\text{d}x\,\text{d}y\,\text{d}z
=
2\cdot\int_{\sqrt{3}/2}^1 \pi\cdot(1-x^2)\,\text{d}x
$$
obtaining the result...
You can also make a trasformation in cylindrical coordinates, like suggested by Zarrax, obtaining
$$
2\cdot\iiint_T \,\text{d}x\,\text{d}y\,\text{d}z
=
2\cdot\iiint_\Omega\rho\,\text{d}\rho\,\text{d}\theta\,\text{d}x
$$
in the set given by the relations
$$
\Omega=\left\{ (\rho,\theta,x)\in\mathbb{R}^3 \Big| \frac{1}{2}\leq\rho\leq\sqrt{1-x^2},\,\frac{\sqrt{3}}{2}\leq x\leq 1, 0\leq\theta\leq 2\pi \right\}, 
$$
arriving at the same conclusions, because
$$
2\cdot\iiint_\Omega \rho \,\text{d}\rho\,\text{d}\theta\,\text{d}x
=
2\cdot\int_{\sqrt{3}/2}^1\int_0^{2\pi}\left[\frac{\rho^2}{2}\right]_0^{\sqrt{1-x^2}}\,\text{d}\theta\,\text{d}x
=
2\pi\cdot\int_{\sqrt{3}/2}^1 (1-x^2)\,\text{d}x\,...
$$
