$$\begin{aligned} &\text { Given equation of the curve is } y=\sqrt{a^2-x^2} \text {. }\\ &\Rightarrow \quad y^2=a^2-x^2 \Rightarrow y^2+x^2=a^2 \end{aligned}$$

$$\begin{aligned} &\therefore \text { Required area of shaded region, }\\ &\begin{aligned} A & =\int_0^a \sqrt{a^2-x^2} d x \\ & =\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a \\ & =\left[0+\frac{a^2}{2} \sin ^{-1}(1)-0-\frac{a^2}{2} \sin ^{-1} 0\right] \\ & =\frac{a^2}{2} \cdot \frac{\pi}{2}=\frac{\pi a^2}{4} \text { sq units } \end{aligned} \end{aligned}$$

$$\begin{aligned} & \text { Given equation of curves are } \quad y=\sqrt{x} \text { and } y=x \text {. } \\ & \Rightarrow \quad x=\sqrt{x} \Rightarrow x^2=x \\ & \Rightarrow \quad x^2-x=0 \Rightarrow x(x-1)=0 \\ & \Rightarrow \quad x=0,1 \end{aligned}$$

$$\begin{aligned} &\therefore \text { Required area of shaded region, }\\ &\begin{aligned} A & =\int_0^1(\sqrt{x}) d x-\int_0^1 x d x \\ & =\left[2 \cdot \frac{x^{3 / 2}}{3}\right]_0^1-\left[\frac{x^2}{2}\right]_0^1 \\ & =\frac{2}{3} \cdot 1-\frac{1}{2}=\frac{2}{3}-\frac{1}{2}=\frac{1}{6} \text { sq units } \end{aligned} \end{aligned}$$

We have, $y=-x^2$ and $x+y+2=0$

$$\begin{array}{lrl} \Rightarrow & -x-2 & =-x^2 \Rightarrow x^2-x-2=0 \\ \Rightarrow & x^2+x-2 x-2 & =0 \Rightarrow x(x+1)-2(x+1)=0 \\ \Rightarrow & (x-2)(x+1) & =0 \Rightarrow x=2,-1 \end{array}$$

$$ \begin{aligned} \therefore \quad \text { Area of shaded region, }A & =\left|\int_{-1}^2\left(-x-2+x^2\right) d x\right|=\left|\int_{-1}^2\left(x^2-x-2\right) d x\right| \\ & =\left|\left[\frac{x^3}{3}-\frac{x^2}{2}-2 x\right]_{-1}^2\right|=\left|\left[\frac{8}{3}-\frac{4}{2}-4+\frac{1}{3}+\frac{1}{2}-2\right]\right| \\ & =\left|\frac{16-12-24+2+3-12}{6}\right|=\left|-\frac{27}{6}\right|=\frac{9}{2} \text { sq units } \end{aligned}$$

Given equation of the curves are $y=\sqrt{x}$ and $x=2 y+3$ in the first quadrant.

$$\begin{aligned} &\text { On solving both the equations for } y \text {, we get }\\ &\begin{array}{r} & y =\sqrt{2 y+3} \\ \Rightarrow & y^2 =2 y+3 \\ \Rightarrow & y^2-2 y-3 =0 \\ \Rightarrow & y^2-3 y+y-3 =0 \\ \Rightarrow & y(y-3)+1(y-3) =0 \\ \Rightarrow & (y+1)(y-3) =0 \\ \Rightarrow & y =-1,3 \end{array} \end{aligned}$$

$$\begin{aligned} &\therefore \text { Required area of shaded region, }\\ &\begin{aligned} A & =\int_0^3\left(2 y+3-y^2\right) d y=\left[\frac{2 y^2}{2}+3 y-\frac{y^3}{3}\right]_0^3 \\ & =\left[\frac{18}{2}+9-9-0\right]=9 \text { sq units } \end{aligned} \end{aligned}$$

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

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

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