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

Marks : 3

Year : 2014

Limit of y are

$y=0$ as straight line

$$y=\sqrt{a^2-x^2}$$

$$x^2+y^2=a^2$$

Which is equation of circle with center at $(0,0)$ and radius a

Limits of x are $x=0$ to $x=a $

On changing the polar co-ordinates,

We take $x=r\cos ɵ y=r\sin ɵ dxdy=rdrd ɵ$

$$x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2$$

Limits of r is $r=0$ to $r=a$ This step will slide from $ɵ=0 $ to $ ɵ=π/2$ to cover whole area.

$\therefore I=\int\limits_{\theta=0}^{\theta=\pi/2}\int\limits^a_{r=0}drd\theta\\ = \int\limits^{\frac \pi2}_0\Bigg[\dfrac {r^4}4\Bigg]_0^ad\theta\\ =\int\limits^{\frac \pi2}_0\dfrac {a^4}4d\theta\\ I=\dfrac {a^4}4[\theta]^{\pi/2}_0\\ =\dfrac \pi4a^4$