The method used is a little sneaky, as it seems to be a Riemann sum in disguise. But it doesn't really feel like calculus, and simply relies on two formulas for series, one for the sum of the integers between 1 and n, and the second for the sum of the squares of the integers:
Consider a hemisphere with radius r, as shown in the figure.
We'll divide it up into horizontal slices, figure the volume of each, and then add up the volumes for all of the slices, yielding the volume of the entire hemisphere. Although the volumes for the individual slices are not quite right (because x isn't the same at the top and bottom of the slice), if we make the slices thin enough, this won't matter.
Cut the sphere into n slices. To make the math simpler, we'll throw one away, considering only n-1 of them. If this bothers you, remember that eventually n will become very large, so that n is nearly equal to n-1.
The width of each slice is r/n. Let y = 0 at the top.
Enumerate the slices with the index k. For the kth slice, y = kr/n. And by the Pythagorean theorem:
The volume of each slice is then:
We need to add up these values for k = 1 to k = n - 1. Take the terms in the brackets one at a time. The first is:
Using the standard formula that the sum of the integers from 1 to n = n(n+1)/2, canceling the 2 and remembering that we're actually summing from 1 to n-1:
Notice that we have just enough n's on the bottom to cancel. For the (n-1) term, dividing by n we get (1 - 1/n) ≈ 1. The entire first term reduces to:
The second term is:
Using the standard formula that the sum of the squares of the integers from 1 to n = n(n+1)(2n+1)/6, and remembering that we're summing from 1 to n-1:
Again, we have just enough n's on the bottom to cancel.
Putting it all together, we have
The complete sphere is twice that:
That was painless!