Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 18 / b

Table of contents
- Solution b

Solution

The domain of integration is the upper half of the ellipse

x^{2} + 4 y^{2} \leq 1

. Reversing the order makes

- 1 \leq x \leq 1

and

0 \leq y \leq \frac{1}{2} 1 - x^{2}

. Thus

\int_{- 1}^{1} \int_{0}^{1 - x^{2} /2} y d y d x = \frac{1}{8} \int_{- 1}^{1} (1 - x^{2}) d x = \frac{1}{6} .

The inner antiderivative is

y^{2} /2

, so the square of the upper limit removes the square root.

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