The remaining step then is to show that the volume is independent of the size of the sphere. We will show that in fact the volume is equal to that of a sphere of radius 5. We will do this by showing that the cross sectional area of the volume in question at each slice is equal to the cross sectional area of the same slice through a sphere of radius 5.
Let the sphere radius be s, and the plug radius be p. The first relation we have is: (1) s^2 = p^2 + 5^2 Let the z axis run down the length of the plug, so that the system is symmetrical around the z axis. Compare the sphere of radius 5 with the (larger) sphere-plug as z goes from -5 to 5. For each z value we will show that the cross section of the two is the same. For the first case, the sphere of radius 5, the cross section is a circle. Call its radius, r. Then we have: (2) r^2 + z^2 = 5^2 The area is pi r^2 and using (2) to substitute for r^2 we get as the cross sectional area of the sphere of radius 5: (3) A_sphere = pi (5^2 - z^2) Now for the larger sphere minus plug. The cross section is an annulus (the space between two circles). The inner radius of the annulus is always p regardless of z, so the area of the inner circle is pi p^2. Substituting (1) above we get: (4) A_inner = pi (s^2 - 5^2) Let the radius of the outer circle at height z be t, and we have (5) t^2 + z^2 = s^2 similarly to (2). The area of this circle is pi t^2 which is: (6) A_outer = pi (s^2 - z^2) The area of the annulus is A_outer - A_inner or: (7) A_annulus = pi (5^2 - z^2) which is the same as (3), as required. Q.E.D. Now let's see you shortcut this proof by "cheating".
