Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 16 / b / iii

Table of contents
- Solution iii

Solution

Use the prism bounds read from the PDF diagram and the divergence from part (ii):

\int_{V} \nabla \cdot F d V = \int_{0}^{2} \int_{0}^{4 - 2 x} \int_{0}^{1} (2 y + 4 x - 8 - 2 z) d z d y d x = \int_{0}^{2} \int_{0}^{4 - 2 x} (2 y + 4 x - 9) d y d x .

A = 4 - 2 x

in the inner evaluation. Since

4 x - 9 = - 2 A - 1

, the inner integral is

A^{2} - (2 A + 1) A = - A^{2} - A

. Therefore

\int_{V} \nabla \cdot F d V = - \int_{0}^{2} [(4 - 2 x)^{2} + (4 - 2 x)] d x = - \frac{32}{3} - 4 = - \frac{44}{3} .

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