POSSIBLE SOLUTION(S)
We consider the problem of a hemisphere of radius 8" as a half of the shape of a rotated circle, x^2 + y^2 = 8^2, of radius 8" about the x axis.
We obtain the surface area of the slice of width a going from x value h to x value h + a.
We do this by rotating an element of arc length Sqrt[1 + f'[x]^2] through a radius f[x] and taking one-half of this amount.
Input :=
va[h_] = 1/2 Integrate[2 Pi f[x] Sqrt[1 + f'[x]^2],
{x,h,h+a}]
Output =
1 2
(-16 h Sqrt[-------] Sqrt[64 - h ] Pi +
2
64 - h
2
2 (a + h) Sqrt[64 - (a + h) ]
-64
Sqrt[--------------] Pi) / 2
2
-64 + (a + h)
In our case we are suggesting that we cut the total width of the 16" diameter hemisphere of bread into 8 equal width pieces of length a = 2".
Thus we compute the surface area from h to h + 2. Here h is in the domain h = -8" to h = 6".
Input :=
party[h_] = va[h]/.{a->2}
Output =
1 2
(-16 h Sqrt[-------] Sqrt[64 - h ] Pi +
2
64 - h
2
2 (2 + h) Sqrt[64 - (2 + h) ]
-64
Sqrt[--------------] Pi) / 2
2
-64 + (2 + h)