Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer) Find the volume common to two circular cylinders, each with radius r, if the axes of the cylinders intersect at right angles. (using disk/washer)

I saw no example of this problem anywhere.. I saw an example how to solve it without calculus but I'm afraid that'll be deductions on my test.. I have the volume of a cylinder and the method of area between two curves but this is out of my reach.... I don't know what I can try...
The main challenge in this problem is to predict the solid itself. Take a look at the image below

As we increase the height $z$ of the intersecting plane the section of the solid is a square with a side $a$. Since the solid is created with two cylinders of radius $r$, $a$ and $z$ are related by $$a^2+z^2=r^2$$
The area of the square is $A=a^2$ or in terms of $z$: $$A(z)=a^2=r^2-z^2$$ The method of washers tells us $$V=8\int_0^rA(z)dz=8\int_0^r(r^2-z^2)dz=8\left(r^3-\frac{r^3}{3}\right)=\frac{16}{3}r^3$$
Note that the coefficient $8$ is required since we are considering a $1/8$th part of the solid.
The volume of a Steinmetz solid $(SS)$, formed by the intersection of $2$ cylinders (radii $r$) at right angles is:

$$V = 4r \cdot r \cdot r \left(\frac{\pi}{2} - \frac{1}{3}\right) = 4r^3\left(\frac{\pi}{2} - \frac{1}{3}\right)$$

As the volume of $SS$ is formed by sections of a cylinder, the answer must contain $\pi$.