Answer
$ \frac{11}{12}\pi$
Work Step by Step
We first rewrite the first equation in the standard form of the equation of a sphere by completing the square:
$x^{2} +y^{2}+z^{2} +4x-2y+4z+5=0$
$x^{2} +4x+4+y^{2} -2y+1+z^{2}+4z+4=-5+4+1+4$
$(x+2)^{2}+(y-1)^{2} +(z+2)^{2}=4$
This is the equation of a sphere with center (-2, 1, -2) and radius 2.
The other sphere has equation
$x^{2}+y^{2} +z^{2}=4$,
which shows that it has center (0, 0, 0) and radius 2.
We are looking for the volume of the solid that lies in both spheres, that is, the volume of the intersection of these two spheres. To find this, it'll be useful to find the distance between the centers:
$\sqrt {(0+2)^{2}+(0-1)^{2} +(0+2)^{2}}$
= $\sqrt {4+1 +4}$
= $\sqrt {9}$ = $3$
Since each sphere has a radius of 2, this means that each sphere reaches into the other 1 unit (if the spheres were only touching at one point, this distance would be 4). Thus the intersection is 1 unit "wide".
If we cut this intersection in half perpendicularly to the segment connecting the centers, we get two identical spherical caps, each of height 0.5.
We can find the volume of these caps using the formula
$V_{c} = \frac{1}{3} \pi h^{2} (3r - h)$
where $h$ is the height of the cap and $r$ is the radius of the sphere.
Since we have two identical caps, the volume of the entire solid will be:
$V = 2 (\frac{1}{3} \pi h^{2} (3r - h))$
$ = 2 (\frac{1}{3} \pi (0.5)^{2} (3(2) - 0.5))$
$ = \frac{2}{3} \pi (0.25) (5.5)$
$ = \frac{11}{12}\pi$
