Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 11 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

r^{2} = x^{2} + y^{2} + z^{2}

, the gradient is

v = 2 e^{r^{2}} (x, y, z)

. Its divergence, equivalently the Laplacian of

ϕ

, is

\nabla \cdot v = (6 + 4 r^{2}) e^{r^{2}} .

On the unit sphere this scalar is the constant

10 e

. The lower hemisphere has surface area

2 π

, so

\int_{S} (\nabla \cdot v) d S = 20 π e .

This is a scalar surface integral of the divergence, not the electric- or vector-field flux integral. There is no normal component to insert.

ii

Solution

The scalar integrand still equals

10 e

throughout the unit sphere, and the upper hemisphere has the same surface area

2 π

. Excluding its equator changes no area. Therefore

\int_{S^{'}} (\nabla \cdot v) d S = 20 π e .

There is no sign reversal between hemispheres because this scalar surface integral has no choice of normal orientation.

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