Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 4

Year : 2015

Here the limit of x is $X=0$ to $x= ∞$ and y is $y=0$ to $y=∞$

On changing to polar co-ordinates

$$X=r \cos ɵ ,y=r \sin ɵ dxdy=r\space drdɵ$$

$\therefore x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta\\ =r^2\\ \therefore I\int\limits\int\limits e^{-r^2}r\space drd\theta $

Here limit of ɵ

$ɵ =0$ to $π$ and limit of r $$ r=0 \space to\space r=∞$$

$\therefore I= \int\limits_0^{\frac \pi2}\int\limits_0^{\infty}re^{-r^2} drd\theta\\ Put \space U=r^2\\ \therefore dU=2rdr\\ \therefore \dfrac {dU}2=rdr\\ When \space r=0 U=0 \\When \space r=∞ U=∞ \\ I=\int\limits_{\frac \pi2}^{\frac \pi2}\int\limits_0^{\infty}e^{-U}\dfrac {dU}2d\theta \\ I= \int\limits_0^{\frac \pi2}\dfrac 12\Bigg[\dfrac {e^{-U}}{-1}\Bigg]_0^{\infty}d\theta\\ = \int\limits_0^{\frac \pi2}\dfrac 12 [-0+1]d\theta\\ =\dfrac 12[\theta]_0^{\frac \pi2}\\ I=\dfrac \pi4 $