 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/13/b/ii/solution

On the loop,

y = t x

, so

d y = t d x + x d t

and

x d y - y d x = x^{2} d t

. The traversal is anticlockwise. The area formula from Green's theorem becomes

A_{loop} = \frac{1}{2} \int_{0}^{\infty} x (t)^{2} d t = \frac{9 a ^{2}}{2} \int_{0}^{\infty} \frac{t ^{2}}{( 1 + t ^{3} ) ^{2}} d t .

u = 1 + t^{3}

to obtain

A_{loop} = \frac{3 a ^{2}}{2} \int_{1}^{\infty} u^{- 2} d u = \frac{3 a ^{2}}{2} .

Although the parameter interval is unbounded, the return to the origin closes the loop and the integral converges.

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