The bowl is in the shape of a truncated hemisphere. For this exercise, R is the large radius and r is the radius of the flat. Since the hemisphere is truncated, the formula for the volume of a sphere cannot be used. The easiest way to find the volume is to use the triple integral with the differential element dz dr dθ

The first thing to do is to find the limits of integration. They are found in reverse order here. dθ will sweep around the circle in the x-y plane, so it will go from 0 to 2π (0 to 360º). dr will sweep from the small radius to the large radius, so it will go from r to R.

The limits for dz are a little different. The equation for a hemisphere is x² + y² + z² = R².

From the Pythagorean Theorem we know that x² + y² = q² (q will be used instead of r for the resultant to avoid confusion with the radius of the flat surface). Substituting this into the first equation gives q²+z²=R².

Solving this for z gives which will be the upper limit. The lower limit is 0.

We also need to know the height of the cylindrical section in the middle. Using h for the height, we know that r² + h² = R², so

Putting everything together gives us:

And after integrating and evaluating dz:

Now a u-substitution is required. Let u = R² - q² and then du = -2q dq.

We have to multiply by -1/2 outside of the integral and -2 inside. (This is the same as multiplying by 1.)

After this multiplication we can substitute du for -2q dq, but we must fix the limits first by substituting the limits for dq into the equation for the u-substitution. For the upper limit u = R² - R² = 0, and for the lower limit u = R² - r²

Now we integrate and evaluate the rest of the equation

This gives the general equation for the volume of a truncated hemisphere. With the amount of manipulation that was necessary a check is in order. In the check R is the large radius and r = 0, thus finding the volume of a hemisphere.

               

This is what we should end up with.

Letting r = 0

This answer checks out correctly, so we have the correct general equation.

To show that there is a difference I measured one of my mixing bowls. The large radius is 5" and the small radius is 2", therfore R = 5 and r = 2

These results show that the flat bottom does make a difference in the volume of the bowl.