Calculate the volume of solid 1° quadrant bounded by: $\displaystyle x^2+y^2=4$ and $\displaystyle x^2+z^2=4$
Answer:
$\displaystyle \frac{16}{3}$
$\displaystyle V = \frac{1}{8} \int_{x = -2}^{x = 2} \int_{y = -\sqrt{4 - x^2}}^{y = + \sqrt{4 - x^2}} \int_{z = -\sqrt{4 - x^2}}^{z = +\sqrt{4 - x^2}} dz \, dy \, dx$.
Which is the same as $\displaystyle V = \int_{x = 0}^{x = 2} \int_{y = 0}^{y = + \sqrt{4 - x^2}} \int_{z = 0}^{z = +\sqrt{4 - x^2}} dz \, dy \, dx$.
Now integrate.