 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017/8/c/solution

The boundary consists of the spherical zone

- a \leq z \leq a

and the inner cylindrical wall. On the sphere,

n = r / R

, so

n \cdot r = R

. Its area element is

R d ϕ d z

\int_{- a}^{a} \int_{0}^{2 π} R^{2} d ϕ d z = 4 πa R^{2} .

On the inner cylinder, the outward normal of the remaining material points towards the hole:

n = - e_{ρ}

. Therefore

n \cdot r = - r_{0}

, and

d S = r_{0} d ϕ d z

- 4 πa r_{0}^{2}

. There are no planar end faces of positive area: the two surfaces meet along the rim circles. Hence

V = \frac{4 πa}{3} (R^{2} - r_{0}^{2}) = \frac{4 π a ^{3}}{3} = \frac{π h ^{3}}{6} .

Using the cylinder's radial normal with the opposite sign would count the hole as extra material and lose the required cancellation.

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