Find out smaller area bounded. between ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $x=a e$, where $a>b$.

Required Area

$=2 \times A$

$=2 \int_{x=a e}^a y d x$

$y= \pm \sqrt[b]{\left(1-\frac{x^2}{a^2}\right)}$

$\quad y=b \sqrt{1-\frac{x^2}{a^2}}=\frac{b}{a} \sqrt{a^2-x^2}$

$=2 \int_{a e}^a \frac{b}{a} \sqrt{a^2-x^2} d x \ =\frac{2 b}{a} \int_{a e}^a \sqrt{a^2-x^2} d x$

$=\frac{2 b}{a}\left[\frac{1}{2} x \sqrt{a^2-x^2}+\frac{1}{2} a^2 \sin ^{-1} \frac{x}{a}\right]_{a e}^{a}$

$=\frac{2 b}{a}\left[0+\frac{1}{2} a^2 \sin ^{-1} 1-\frac{1}{2} a e \sqrt{a^2-a^2 e^2}-\frac{1}{2} a^2 \sin ^{-1} e\right]$

$=\frac{2 b}{a}\left[\frac{\pi a^2}{4}-\frac{1}{2} a e b-\frac{1}{2} a^2 \sin^{-1} c\right]$

Case $2 \quad f(x) \geqslant g(x), x \in[0, c]$

$g(x) \geqslant f(x), \quad x \in[c, b]$

Required Area

$\int_ a^c[f(x)-g(x)] d x$ + $\int_c^b[g(x)-f(x)] d x$

$\begin{aligned}& \int_{x=0}^{x=\frac{1}{2}}\left(\sqrt{2 x}-4 x^2\right) d x \end{aligned}$

$\begin{aligned}& =\left.\left.\sqrt{2} \frac{x^{3 / 2}}{3 / 2}\right|_0 ^{\frac{1}{2}} 4 \frac{x^3}{3}\right|_0 ^{1 / 2} \end{aligned}$

$\begin{aligned}& =\frac{2 \sqrt{2}}{3} \frac{1}{2} \frac{1}{\sqrt{2}}-\frac{4}{3} \times \frac{1}{8} \end{aligned}$

$\begin{aligned}& =\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\end{aligned}$

$\begin{array}{r}y^2=2 x, y=4 x^2 \ 16 x^4=2 x\left|\begin{array}{l}x=0 \ x\left(8 x^3-x\right)=0\end{array}\right| x=\frac{1}{2}\end{array}$

Find out area of region which is out side parabola $y^2=4 x$ and inside the circle $x^2+y^2 \equiv 8 x,(x-4)^2+y^2 \equiv 16$ $x^2+4 x=8 x$ $x^2=4 x$ $x^2=4 x$ $x(x-4)=0$ $x=0, x=4$

Solution:Required Area is symmetrical about $x$-axis. Required area $=2 A$ axis,

$\int_{x=0}^{x=4} \frac{\left(\sqrt{16-(x-4)^2}-\sqrt{4 x}\right) d x}{x-4=t}$

$y=\sqrt{16-(x-4)^2}$

$y= \pm \sqrt{4 x}$ $= \int_{-4}^0 \sqrt{16-t^2} d t-2 \int_0^4 \sqrt{x} d x$

$x=0 \quad x \quad y= \pm \sqrt{4 x} =\int_{-4}^0 \sqrt{16-t^2} d t-2 \int_0^4 \sqrt{x} d x$

$=\left[\frac{1}{2} t \sqrt{16-t^2}+\frac{1}{2} 16 {\sin}^{-1} \frac{t}{4}\right]^0_{-4}$

• $- 2\frac{x^{3 / 2}}{3 / 2}]_0^4$

• $=0+0-0-8 \sin ^{-1}(-1)-\frac{4}{3} \times(4)^{3 / 2}+0$

• $=4 \pi-\frac{4}{3} \times 8=4 \pi-\frac{32}{3} . \ \text { Req in area }=8 \pi-64 / 8 .$

Area bounded between circle $x^2+y^2=4$ and $(x-2)^2+y^2=4$.

Solution:

Elementary Area

$=\left[\sqrt{4-y^2}-\left(2-\sqrt{4-y^2}\right)\right] d y$

$=\left[2 \sqrt{4-y^2}-2\right] d y$

$x-2= \pm \sqrt{4-y^2}$

$x=2 \pm \sqrt{4-y^2}$

$x=2 \pm \sqrt{4-y^2} x^2+y^2=4$

$x= \pm \sqrt{4-y^2}$

$\int_{y=-}^{\left(2 \sqrt{4-y^2}-2\right) d y}$

$(x-2)^2+x-x^2=4 x^2-4 x+4-x^2=0 x=1 y= \pm \sqrt{3}$

$\int_{y=-\sqrt{3}}^{\left(\frac{1}{3} \sqrt{4-y^2}-2\right) d y}$

${c}(x-2)^2+x^2-x^2=4 x^2-4 x+4-x^2=0 x=1 y= \pm \sqrt{3} =2$

$\int_0^{\sqrt{3}}\left(2 \sqrt{4-y^2}-2\right) d y =4$

$\int_0^{\sqrt{3}}\left(\frac{\sqrt{4-y^2}}{2}-1\right) d y$.

$=4\left[\frac{1}{2} y \sqrt{4-y^2}+\frac{1}{2} 4 \sin ^{-1} \frac{y}{2}-y\right]_0^{\sqrt{3}}$

$=4 \int_0^{\sqrt{3}}\left({\sqrt{4-y^2}}^2-1\right) d y .$

$= 4\left[\frac{1}{2} y \sqrt{4-y^2}+\frac{1}{2} 4 \sin ^{-1} \frac{y}{2}-y\right]_0^{\sqrt{3}}$

$= 4\left[\frac{1}{2} \sqrt{3}+2 \sin ^{-1} \frac{\sqrt{3}}{2}-\sqrt{3}\right]$

$= 4\left[2 \frac{\pi}{3}-\frac{1}{2} \sqrt{3}\right] \ = \frac{8 \pi}{3}-2 \sqrt{3}$

Ex. Area bounded between $x$-axis, $\sin x$ and $x=-\frac{\pi}{2}$ is $x=\frac{3 \pi}{2}$

Required Area.

$\text { Required Area }$

$A_1=\int_{-\pi / 2}^0 \sin x d x=-\left.\cos x\right|_{-\pi / 2} ^0=-1$

$A_2=\int_0^\pi \sin x d x=-\left.\cos x\right|_0 ^\pi=1-(-1)=2$

$A_3=\int_\pi^{3 \pi / 2} \sin x d x=-\left.\cos x\right|_\pi ^{3 \pi / 2}=-1$

Required Area.

$=\left|A_1\right|+A_2+\left|A_3\right|$

$A_1=\int_{-\pi / 2}^0 \sin x d x=-\left.\cos x\right|_{-\pi / 2} ^0=-1$

$A_2=\int_0^{\pi / 2} \sin x d x=-\left.\cos x\right|_0 ^\pi=1-(-1)=2$

$A_3=\int_\pi^{3 \pi / 2} \sin x d x=-\left.\cos x\right|_\pi ^{3 \pi / 2}=-$

Ex. Find out area bounded by $|x|+|y|=1$.

$\int_0^1(1-x-(x-1)) d x$

$+\int_{-1}^0(1+x-(-x-2)) d x=1, \quad x-y=1,-x-y=1 .$

$.=\int_0^1 2(1-x) d x+\int_{-1}^0 2(1+x)$

$d x=\left.2 \frac{(1-x)^2}{-2}\right|_0$

$^1+\frac{(1+x)^2}{2}]_1^0$

$=-2 \times\left(-\frac{1}{2}\right)+2 \times \frac{1}{2}=1+1=2$

Example. Find out area bounded between this Curves.

$y^2=4 a x$

$y=\underline{m} x \text {. }$

$\underline{a}>0, m>0 .$