 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-1/12/solution

The region lies in the first quadrant, inside the unit circle and above the positive branch of

x y = 1/4

x y = (r^{2} /2) sin 2 θ

, so its radial bounds are

(2 sin 2 θ)^{- 1/2} \leq r \leq 1

. Such an interval exists when

sin 2 θ \geq 1/2

, hence

\frac{π}{12} \leq θ \leq \frac{5 π}{12}, \frac{1}{2 sin 2 θ} \leq r \leq 1.

At the intersections with the circle,

r = 1

and

sin 2 θ = 1/2

, giving exactly the two stated polar angles. Their Cartesian coordinates are

(cos (π /12), sin (π /12))

and the swapped pair.

d x d y = r d r d θ

, while

x y (x^{2} + y^{2}) = r^{4} sin 2 θ /2

I = \int_{π /12}^{5 π /12} \int_{(2 s i n 2 θ)^{- 1/2}}^{1} \frac{r ^{5}}{2} sin 2 θ d r d θ = \int_{π /12}^{5 π /12} (\frac{s i n 2 θ}{12} - \frac{c s c ^{2} ( 2 θ )}{96}) d θ .

- cos 2 θ /24 + cot 2 θ /192

. Evaluating the bounds gives

I = \frac{3}{24} - \frac{3}{96} = \frac{3}{32} .

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