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

Marks : 3

Year : 2014

The region of integration is bounded by $y = 0$ and $y = \sqrt{2ax-x^2}$

$Y^2=2ax-x^2 \\ X^2+y^2-2ax=0$

Adding $+a^2$ both side we get

$X^2-2ax+a^2+y^2=a^2 \\ (x-a)^2+y^2=a^2$

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

The region of integration is also bounded by $x = 0$ to $x = 2a$

If we put $x = r\cosθ$ and $y = r\sinθ$

For inner limits

$y=\sqrt{2ax-x^2} \\ y^2={2ax-x^2} \\ x^2+y^2=2ax \\ x^2=2a*\cos\theta \\ \therefore y=2a\cos\theta \\ y=0 \\ x\sin\theta=0 \\ \therefore x=0$

For outer limits

Here the radial strip will vary from $0 \ to \ \dfrac \pi2$

dx dy is replaced by r dr dθ

Our integration becomes

$=\int\limits_0^\frac{\pi}{2}\int\limits_0^{2a\cos\theta}\frac {r\cos\theta}{\sqrt{r^2}}rdrd\theta \\ =\int\limits_0^\frac{\pi}{2}\int\limits_0^{2a\cos\theta}r\cos\theta \space\space drd\theta \\ =\int\limits_0^\frac {\pi}{2}\Bigg[\frac {r^2}2\cos\theta\Bigg]_0^{2a\cos\theta} \\ =\int\limits_0^\frac {\pi}{2}\Bigg[\frac {4a^2}2\cos^2\theta\times\cos\theta\Bigg]d\theta \\ =2a^2\int\limits_0^\frac {\pi}{2}(1-\sin^2\theta)\times\cos\theta d\theta \\ I=2a^2\Bigg[\int\limits_0^\frac{\pi}{2}\cos\theta \space d\theta- \int\limits^\dfrac \\ \pi2_0\sin^2\theta\cos\theta\space d\theta\Bigg] \\ I_1=\int\limits_0^\frac{\pi}{2}\cos\theta\space d\theta=[\sin\theta]_0^{\pi/2}=1 \\ I_2=\int\limits_0^\frac{\pi}{2}\sin^2\theta\cos\theta\space d\theta$

Put $\sin\theta=t \\ \cos\theta\space d\theta=dt \\ when \ \theta=0\space \space t=0$

$\theta=\pi/2 \space \space t=1 \\ =\int\limits_0^1t^2 dt$

$=\Bigg[\frac{t^3}{3}\Bigg]_0^1 \\ =\frac{1}{3}$

Resubstituting value of integration 1 and 2 we get

$I=2a^2\Bigg[1-\dfrac 13\Bigg] \\ I=\dfrac {4a^2}3$