 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/1/solution

For a continuously differentiable vector field on a volume

V

with piecewise smooth boundary

S

\iint_{S} A \cdot \hat{n} d S = ∭_{V} \nabla \cdot A d V,

where

\hat{n}

is the outward unit normal. For the polynomial field here,

\nabla \cdot A = \partial_{x} (4 x z) + \partial_{y} (- y^{2}) + \partial_{z} (yz) = 4 z - y .

Therefore the outward flux through the unit cube is

I = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (4 z - y) d x d y d z = 2 - \frac{1}{2} = \frac{3}{2} .

As a direct boundary check, the

x = 1

face contributes

2

, the

y = 1

face contributes

- 1

, and the

z = 1

face contributes

1/2

. All three opposite faces contribute zero, reproducing the result and fixing the orientation sign.

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