Let r bet the radius of both cylinders and their axes are both perpendicular and intersecting. There is a common sphere enveloped by the 2 intersecting cylinders. If we look at the planar section which passes through both cylinder axes we can see a square area common to both cylinders. This square has an area of (2.r.2.r). We can also see a circle which is produced as a result of the plane passing through the enveloped common sphere. This circle has an area of (Pi.r.r).
If the plane is moved away from the cylinders axes, the intersection surface will reduce until the plane exits the cylinders completely. During this transition an ever changing common surface area is produced, but it will always consist of a square with a circle inside it which is tangential to that square at the middle of each side of the square. The ratio of the square to the circle is (4.r.r)/(pi.r.r)= 4/pi.
We know the volume of the common sphere is (4.Pi.r.r.r/3),So by using a similar shape approach, we can deduce that the common volume of the intersecting cylinders will be (4.Pi.r.r.r/3).(4/Pi)= 16.r.r.r/3 There is no Pi in the answer. hope that's useful, and no Calculus either. Lol.