$$\begin{aligned} \text{We have,}\quad y^2=9 x \text { and } y & =3 x \\ \Rightarrow\quad (3 x)^2 & =9 x \\ \Rightarrow\quad 9 x^2-9 x & =0 \\ \Rightarrow\quad 9 x(x-1) & =0 \\ \Rightarrow\quad x & =1,0 \end{aligned}$$

$$\begin{aligned} &\therefore\quad \text { Required area, }\\ &\begin{aligned} A & =\int_0^1 \sqrt{9 x} d x-\int_0^1 3 x d x \\ & =3 \int_0^1 x^{1 / 2} d x-3 \int_0^1 x d x \\ & =3\left[\frac{x^{3 / 2}}{3 / 2}\right]_0^1-3\left[\frac{x^2}{2}\right]_0^1 \\ & =3\left(\frac{2}{3}-0\right)-3\left(\frac{1}{2}-0\right) \\ & =2-\frac{3}{2}=\frac{1}{2} \text { sq units } \end{aligned} \end{aligned}$$

$$\begin{array}{lrl} \text { We have, } & y^2=2 p x \text { and } x^2 & =2 p y \\ \therefore & y & =\sqrt{2 p x} \end{array}$$

$$ \begin{array}{rl} \Rightarrow & x^2=2 p \cdot \sqrt{2 p x} \\ \Rightarrow & x^4=4 p^2 \cdot(2 p x) \\ \Rightarrow & x^4=8 p^3 x \\ \Rightarrow & x^4-8 p^3 x=0 \\ \Rightarrow & x^3\left(x-8 p^3\right)=0 \\ \Rightarrow & x=0,2 p \end{array}$$

$$\begin{aligned} \therefore \quad \text { Required area } & =\int_0^{2 p} \sqrt{2 p x} d x-\int_0^{2 p} \frac{x^2}{2 p} d x \\ & =\sqrt{2 p} \int_0^{2 p} x^{1 / 2} d x-\frac{1}{2 p} \int_0^{2 p} x^2 d x \\ & =\sqrt{2 p}\left[\frac{2(x)^{3 / 2}}{3}\right]_0^{2 p}-\frac{1}{2 p}\left[\frac{x^3}{3}\right]_0^{2 p} \\ & =\sqrt{2 p}\left[\frac{2}{3} \cdot(2 p)^{3 / 2}-0\right]-\frac{1}{2 p}\left[\frac{1}{3}(2 p)^3-0\right] \\ & =\sqrt{2 p}\left(\frac{2}{3} \cdot 2 \sqrt{2} p^{3 / 2}\right)-\frac{1}{2 p}\left(\frac{1}{3} 8 p^3\right) \\ & =\sqrt{2} p\left(\frac{4 \sqrt{2}}{3} p^{3 / 2}\right)-\frac{1}{2 p}\left(\frac{8}{3} p^3\right) \\ & =\frac{4 \sqrt{2}}{3} \cdot \sqrt{2} p^2-\frac{8}{6} p^2 \\ & =\frac{(16-8) p^2}{6}=\frac{8 p^2}{6} \\ & =\frac{4 p^2}{3} \text { sq units } \end{aligned}$$

We have, $y=x^3, y=x+6$ and $x=0$

$$\begin{array}{lr} \therefore & x^3=x+6 \\ \Rightarrow & x^3-x=6 \\ \Rightarrow & x^3-x-6=0 \\ \Rightarrow & x^2(x-2)+2 x(x-2)+3(x-2)=0 \\ \Rightarrow & (x-2)\left(x^2+2 x+3\right)=0 \\ \Rightarrow & x=2, \text { with two imaginary points } \end{array}$$

$$\begin{aligned} \therefore \text { Required area of shaded region } & =\int_0^2\left(x+6-x^3\right) d x \\ & =\left[\frac{x^2}{2}+6 x-\frac{x^4}{4}\right]_0^2 \\ & =\left[\frac{4}{2}+12-\frac{16}{4}-0\right] \\ & =[2+12-4]=10 \text { sq units } \end{aligned}$$

Given equation of curves are

$$\begin{aligned} & y^2=4 x \quad\text{.... (i)}\\ \text{and}\quad & x^2=4 y\quad\text{.... (ii)} \end{aligned}$$

$\Rightarrow \quad\left(\frac{x^2}{4}\right)^2=4 x$

$$\begin{array}{rr} \Rightarrow & \frac{x^4}{4 \cdot 4}=4 x \\ \Rightarrow & x^4=64 x \\ \Rightarrow & x^4-64 x=0 \\ \Rightarrow & x\left(x^3-4^3\right)=0 \\ \Rightarrow & x=4,0 \end{array}$$

$$\begin{aligned} \therefore \text { Area of shaded region, }A & =\int_0^4\left(\sqrt{4 x}-\frac{x^2}{4}\right) d x \\ & =\int_0^4\left(2 \sqrt{x}-\frac{x^2}{4}\right) d x=\left[\frac{2 x^{3 / 2} \cdot 2}{3}-\frac{1}{4} \cdot \frac{x^3}{3}\right]_0^4 \\ & =\frac{2 \cdot 2}{3} \cdot 8-\frac{1}{4} \cdot \frac{64}{3}-0=\frac{32}{3}-\frac{16}{3}=\frac{16}{3} \text { sq units } \end{aligned}$$

We have, $y^2=9 x$ and $y=x$

$$\begin{array}{rr} \Rightarrow & x^2=9 x \\ \Rightarrow & x^2-9 x=0 \\ \Rightarrow & x(x-9)=0 \\ \Rightarrow & x \end{array}=0,9$$

$$\begin{aligned} &\therefore \text { Area of shaded region, }\\ &\begin{aligned} A & =\int_0^9(\sqrt{9 x}-x) d x=\int_0^9 3 x^{1 / 2} d x-\int_0^9 x d x \\ & =\left[3 \cdot \frac{x^{3 / 2}}{3} \cdot 2\right]_0^9-\left[\frac{x^2}{2}\right]_0^9 \\ & =\left[\frac{3 \cdot 3^{\frac{3}{2} \times 2}}{3} \cdot 2-0\right]-\left[\frac{81}{2}-0\right] \\ & =54-\frac{81}{2}=\frac{108-81}{2}=\frac{27}{2} \text { sq units } \end{aligned} \end{aligned}$$