Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 8 / b

Table of contents
- Solution b

Solution

In first-quadrant polar coordinates,

x = r cos θ

,

y = r sin θ

, and the Jacobian determinant contributes

r

. The integrand and area element together become

r^{2} e^{- r^{2}} sin θ cos^{2} θ d r d θ

. Hence

I = (\int_{0}^{π /2} sin θ cos^{2} θ d θ) (\int_{0}^{\infty} r^{2} e^{- r^{2}} d r) = \frac{1}{3} \frac{π}{4} = \frac{π}{12} .

The radial integral follows by integration by parts from the Gaussian integral. The isolated undefined ratio at the origin does not affect this convergent double integral.

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