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

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

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

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

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

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

Given equations of lines are

$$\begin{gathered} y=4 x+5 \quad\text{.... (i)}\\ y=5-x \quad\text{.... (ii)}\\ \text{and}\quad 4 y=x+5 \quad \text{.... (iii)} \end{gathered}$$

On solving Eqs. (i) and (ii), we get

$$\begin{aligned} 4 x+5 & =5-x \\ \Rightarrow x & =0 \end{aligned}$$

On solving Eqs. (i) and (iii), we get

$$ \begin{aligned} 4(4 x+5) & =x+5 \\ \Rightarrow\quad 16 x+20 & =x+5 \\ \Rightarrow\quad 15 x & =-15 \\ \Rightarrow\quad x & =-1 \end{aligned}$$

$$\begin{aligned} &\text { On solving Eqs. (ii) and (iii), we get }\\ &\begin{array}{r} 4(5-x)=x+5 \\ \Rightarrow 20-4 x=x+5 \end{array} \end{aligned}$$

On solving Eqs. (i) and (iii), we get

$$\begin{aligned} &\text { On solving Eqs. (ii) and (iii), we get }\\ &\begin{array}{r} 4(5-x)=x+5 \\ \Rightarrow\quad 20-4 x=x+5 \end{array} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad x=3\\ & \therefore \text { Required area }=\int_{-1}^0(4 x+5) d x+\int_0^3(5-x) d x-\frac{1}{4} \int_{-1}^3(x+5) d x \\ &=\left[\frac{4 x^2}{2}+5 x\right]_{-1}^0+\left[5 x-\frac{x^2}{2}\right]_0^3-\frac{1}{4}\left[\frac{x^2}{2}+5 x\right]_{-1}^3 \\ &=[0-2+5]+\left[15-\frac{9}{2}-0\right]-\frac{1}{4}\left[\frac{9}{2}+15-\frac{1}{2}+5\right] \\ &=3+\frac{21}{2}-\frac{1}{4} \cdot 24 \\ &=-3+\frac{21}{2}=\frac{15}{2} \text { sq units } \end{aligned}$$

$$\begin{aligned} \text { Required area of shaded region } & =\int_0^{2 \pi} 2 \cos x d x \\ & =\int_0^{\pi / 2} 2 \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} 2 \cos x d x\right|+\int_{3 \pi / 2}^{2 \pi} 2 \cos x d x \end{aligned}$$

$$\begin{aligned} & =2[\sin x]_0^{\pi / 2}+\left|2(\sin x)_{\pi / 2}^{3 \pi / 2}\right|+2[\sin x]_{3 \pi / 2}^{2 \pi} \\ & =2+4+2=8 \text { sq units } \end{aligned}$$

We have, $y=1+|x+1|, x=-3, x=3$ and $y=0$

$\because\quad y=\left\{\begin{array}{cc}-x, & \text { if } x<-1 \\ x+2, & \text { if } x \geq-1\end{array}\right.$

$$\begin{aligned} \therefore \quad \text { Area of shaded region, } A & =\int_{-3}^{-1}-x d x+\int_{-1}^3(x+2) d x \\ & =-\left[\frac{x^2}{2}\right]_{-3}^{-1}+\left[\frac{x^2}{2}+2 x\right]_{-1}^3 \\ & =-\left[\frac{1}{2}-\frac{9}{2}\right]+\left[\frac{9}{2}+6-\frac{1}{2}+2\right] \\ & =-[-4]+[8+4] \\ & =12+4=16 \text { sq units } \end{aligned}$$