We have, $y^2=2 x$ and $x^2+y^2=4 x$

$$\begin{array}{lr} \Rightarrow & x^2+2 x=4 x \\ \Rightarrow & x^2-2 x=0 \\ \Rightarrow & x(x-2)=0 \\ \Rightarrow & x=0,2 \\ \text { Also, } & x^2+y^2=4 x \\ \Rightarrow & x^2-4 x=-y^2 \\ \Rightarrow & x^2-4 x+4=-y^2+4 \\ \Rightarrow & (x-2)^2-2^2=-y^2 \end{array}$$

$$\begin{aligned} \therefore \text { Required area } & =2 \cdot \int_0^2\left[\sqrt{2^2-(x-2)^2}-\sqrt{2 x}\right] d x \\ & =2\left[\left[\frac{x-2}{2} \cdot \sqrt{2^2-(x-2)^2}+\frac{2^2}{2} \sin ^{-1}\left(\frac{x-2}{2}\right)\right]_0^2-\left[\sqrt{2} \cdot \frac{x^{3 / 2}}{3 / 2}\right]_0^2\right] \\ & =2\left[\left(0+0-1 \cdot 0+2 \cdot \frac{\pi}{2}\right)-\frac{2 \sqrt{2}}{3}\left(2^{3 / 2}-0\right)\right] \\ & =\frac{4 \pi}{2}-\frac{8 \cdot 2}{3}=2 \pi-\frac{16}{3}=2\left(\pi-\frac{8}{3}\right) \text { sq units } \end{aligned}$$

$$\begin{aligned} \text { Required area }=\int_0^{2 \pi} \sin x d x & =\int_0^\pi \sin x d x+\left|\int_\pi^{2 \pi} \sin x d x\right| \\ & =-[\cos x]_0^\pi+\left|[-\cos x]_\pi^{2 \pi}\right| \\ & =-[\cos \pi-\cos 0]+|-[\cos 2 \pi-\cos \pi]| \end{aligned}$$

$$\begin{aligned} & =-[-1-1]+|-(1+1)| \\ & =2+2=4 \text { sq units } \end{aligned}$$

Let we have the vertices of a $\triangle A B C$ as $A(-1,1), B(0,5)$ and $C(3,2)$.

$\therefore \quad$ Equation of $A B$ is $y-1=\left(\frac{5-1}{0+1}\right)(x+1)$

$$ \begin{array}{lr} \Rightarrow & y-1=4 x+4 \\ \Rightarrow & y=4 x+5\quad\text{.... (i)} \end{array}$$

and $\quad$ equation of $B C$ is $y-5=\left(\frac{2-5}{3-0}\right)(x-0)$

$$\begin{array}{lr} \Rightarrow & y-5=\frac{-3}{3}(x) \\ \Rightarrow & y=5-x\quad\text{.... (ii)} \end{array}$$

Similarly, $\quad$ equation of $A C$ is $y-1=\left(\frac{2-1}{3+1}\right)(x+1)$

$$\begin{array}{lc} \Rightarrow & y-1=\frac{1}{4}(x+1) \\ \Rightarrow & 4 y=x+5\quad\text{.... (iii)} \end{array}$$

$\therefore \quad$ Area of shaded region $=\int_{-1}^0\left(y_1-y_2\right) d x+\int_0^3\left(y_1-y_2\right) d x$

$$\begin{aligned} & =\int_{-1}^0\left[4 x+5-\frac{x+5}{4}\right] d x+\int_0^3\left[5-x-\frac{x+5}{4}\right] d x \\ & =\left[\frac{4 x^2}{2}+5 x-\frac{x^2}{8}-\frac{5 x}{4}\right]_{-1}^0+\left[5 x-\frac{x^2}{2}-\frac{x^2}{8}-\frac{5 x}{4}\right]_0^3 \\ & =\left[0-\left(4 \cdot \frac{1}{2}+5(-1)-\frac{1}{8}+\frac{5}{4}\right)\right]+\left[\left(15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4}\right)-0\right] \\ & =\left[-2+5+\frac{1}{8}-\frac{5}{4}+15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4}\right] \\ & =18+\left(\frac{1-10-36-9-30}{8}\right) \\ & =18+\left(-\frac{84}{8}\right)=18-\frac{21}{2}=\frac{15}{2} \text { sq units } \end{aligned}$$

We have, $$y^2=6 a x \text { and } x^2+y^2=16 a^2$$

$$\begin{aligned} \Rightarrow \quad & x^2+6 a x =16 a^2 \\ \Rightarrow \quad & x^2+6 \mathrm{a} x-16 \mathrm{a}^2 =0 \\ \Rightarrow \quad & x^2+8 a x-2 \mathrm{a} x-16 \mathrm{a}^2 =0 \\ \Rightarrow \quad & x(x+8 \mathrm{a})-2 \mathrm{a}(x+8 \mathrm{a}) =0 \\ \Rightarrow \quad & (x-2 \mathrm{a})(x+8 \mathrm{a}) =0 \\ \Rightarrow \quad & x =2 \mathrm{a},-8 \mathrm{a} \end{aligned}$$

$$\begin{aligned} & \therefore \text { Area of required region }=2\left[\int_0^{2 a} \sqrt{6 a x} d x+\int_{2 a}^{4 a} \sqrt{(4 a)^2-x^2} d x\right] \\ & \qquad \\ & \quad=2\left[\int_0^{2 a} \sqrt{6 a} x^{1 / 2} d x+\int_{2 a}^{4 a} \sqrt{(4 a)^2-x^2} d x\right] \\ & \\ & =2\left[\sqrt{6 a}\left[\frac{x^{3 / 2}}{3 / 2}\right]_0^{2 a}+\left(\frac{x}{2} \sqrt{(4 a)^2-x^2}+\frac{(4 a)^2}{2} \sin ^{-1} \frac{x}{4 a}\right)_{2 a}^{4 a}\right] \\ & \\ & =2\left[\sqrt{6 a} \cdot \frac{2}{3}\left((2 a)^{3 / 2}-0\right)+\frac{4 a}{2} \cdot 0+\frac{16 a^2}{2} \cdot \frac{\pi}{2}-\frac{2 a}{2} \sqrt{16 a^2-4 a^2}-\frac{16 a^2}{2} \cdot \sin ^{-1} \frac{2 a}{4 a}\right] \\ & \\ & =2\left[\sqrt{6 a} \frac{2}{3} \cdot 2 \sqrt{2} a^{3 / 2}+0+4 \pi a^2-\frac{2 a}{2} \cdot 2 \sqrt{3} a-8 a^2 \cdot \frac{\pi}{6}\right] \\ & \\ & =2\left[\sqrt{12} \cdot \frac{4}{3} a^2+4 \pi a^2-2 \sqrt{3} a^2-\frac{4 a^2 \pi}{3}\right] \\ & \\ & =2\left[\frac{8 \sqrt{3} a^2+12 \pi a^2-6 \sqrt{3} a^2-4 a^2 \pi}{3}\right] \\ & \\ & =\frac{2}{3} a^2[8 \sqrt{3}+12 \pi-6 \sqrt{3}-4 \pi] \\ & \\ & =\frac{2}{3} a^2[2 \sqrt{3}+8 \pi]=\frac{4}{3} a^2[\sqrt{3}+4 \pi] \end{aligned}$$

We have, $$x+2 y=2\quad\text{.... (i)}$$

and

$$\begin{array}{r} y-x=1 \quad\text{.... (ii)}\\ 2 x+y=7\quad\text{.... (iii)} \end{array}$$

$$\begin{aligned} &\text { On solving Eqs. (i) and (ii), we get }\\ &y-(2-2 y)=1 \Rightarrow 3 y-2=1 \Rightarrow y=1 \end{aligned}$$

$$\begin{aligned} &\text { On solving Eqs. (ii) and (iii), we get }\\ &\begin{array}{rr} \Rightarrow & 2 y-2+y=7 \\ \Rightarrow & y=3 \end{array} \end{aligned}$$

$$\begin{aligned} &\text { On solving Eqs. (i) and (iii), we get }\\ &\begin{array}{lr} & 2(2-2 y)+y=7 \\ \Rightarrow & 4-4 y+y=7 \\ \Rightarrow & -3 y=3 \\ \Rightarrow & y=-1 \end{array} \end{aligned}$$

$$\begin{aligned} \therefore \text { Required area } & =\int_{-1}^1(2-2 y) d y+\int_{-1}^3 \frac{(7-y)}{2} d y-\int_1^3(y-1) d y \\ & =\left[-2 y+\frac{2 y^2}{2}\right]_{-1}^1+\left[\frac{7 y}{2}-\frac{y^2}{2 \cdot 2}\right]_{-1}^3-\left[\frac{y^2}{2}-y\right]_1^3 \\ & =\left[-2+\frac{2}{2}-2-\frac{2}{2}\right]+\left[\frac{21}{2}-\frac{9}{4}+\frac{7}{2}+\frac{1}{4}\right]-\left[\frac{9}{2}-3-\frac{1}{2}+1\right] \\ & =[-4]+\left[\frac{42-9+14+1}{4}\right]-\left[\frac{9-6-1+2}{2}\right] \\ & =-4+12-2=6 \text { sq units } \end{aligned}$$