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

Table of contents
- Solution 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.

 Ancestors (13)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version