Find the volume of the body bounded by $x^2 + y^2 + z^2 = 4$ and $x^2 + y^2 = 1$ 
Find the volume of the body bounded by $x^2 + y^2 + z^2 = 4$ and $x^2 + y^2 = 1$.

This is the last subject in our syllabus and I am afraid that my professor had not any time to teach it before the end of the semester, and left us wondering in the land of calculating body volumes in $3D$. I know this can be done by double integral, but I would be happy if someone showed me direction into solving such questions.

The volume can be found as

$$V=\int\int\int dV =  \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}dzdydx .$$

Or just, you can use the double integral as

$$ V = \int\int (z_2-z_1)dA, $$

where $z_1=z_1(x,y)$ and $z_2=z_2(x,y)$ and $dA=dydx$. In your case
$$ z_1= -\sqrt{4-x^2-y^2},\quad z_2= \sqrt{4-x^2-y^2} .$$
Here is a plot, so you can see what's going on
 
