Interpret the curves

The given curves are: $y^2=8x$ and $y=\sqrt{2}\,x$

The parabola can be written as: $x=\frac{y^2}{8}$

The line can be written as: $x=\frac{y}{\sqrt{2}}$

Find points of intersection of the parabola and the line

Substitute $x=\dfrac{y}{\sqrt{2}}$ into $y^2=8x$: $y^2=8\cdot \frac{y}{\sqrt{2}}=4\sqrt{2}\,y$

So, $y(y-4\sqrt{2})=0$

Hence the intersections are at: $y=0 \quad \text{or} \quad y=4\sqrt{2}$

Corresponding $x$-coordinates are:

• for $y=0$, $x=0$
• for $y=4\sqrt{2}$, $x=\dfrac{4\sqrt{2}}{\sqrt{2}}=4$

So intersection points are: $(0,0),\ (4,4\sqrt{2})$

Understand the triangle

The triangle is formed by: $y=\sqrt{2}x, \quad x=1, \quad y=2\sqrt{2}$

Its vertices are obtained from pairwise intersections:

• $x=1$ and $y=\sqrt{2}x \Rightarrow (1,\sqrt{2})$
• $x=1$ and $y=2\sqrt{2} \Rightarrow (1,2\sqrt{2})$
• $y=2\sqrt{2}$ and $y=\sqrt{2}x \Rightarrow x=2 \Rightarrow (2,2\sqrt{2})$

So this triangle lies inside the region enclosed by the parabola and the line.

Area enclosed by the parabola and the line

It is easier to integrate with respect to $y$.

For $0\le y\le 4\sqrt{2}$:

• right curve: line $x=\dfrac{y}{\sqrt{2}}$
• left curve: parabola $x=\dfrac{y^2}{8}$

Therefore, $A_{\text{total}}=\int_0^{4\sqrt{2}}\left(\frac{y}{\sqrt{2}}-\frac{y^2}{8}\right)dy$

Compute: $\int \frac{y}{\sqrt{2}}dy=\frac{y^2}{2\sqrt{2}}, \qquad \int \frac{y^2}{8}dy=\frac{y^3}{24}$

So, $A_{\text{total}}=\left[\frac{y^2}{2\sqrt{2}}-\frac{y^3}{24}\right]_0^{4\sqrt{2}}$

Now, $(4\sqrt{2})^2=32, \qquad (4\sqrt{2})^3=128\sqrt{2}$

Hence, $A_{\text{total}}=\frac{32}{2\sqrt{2}}-\frac{128\sqrt{2}}{24}$ $=\frac{16}{\sqrt{2}}-\frac{16\sqrt{2}}{3}$ $=8\sqrt{2}-\frac{16\sqrt{2}}{3}$ $=\frac{8\sqrt{2}}{3}$

Area of the triangle

The triangle has:

• vertical side from $(1,\sqrt{2})$ to $(1,2\sqrt{2})$, so height $=2\sqrt{2}-\sqrt{2}=\sqrt{2}$
• horizontal distance from $(1,2\sqrt{2})$ to $(2,2\sqrt{2})$, so base $=2-1=1$

Therefore, $A_{\triangle}=\frac{1}{2}\times 1\times \sqrt{2}=\frac{\sqrt{2}}{2}$

Required area

The asked area is the part enclosed by the parabola and the line that lies outside this triangle: $A=A_{\text{total}}-A_{\triangle}$

$A=\frac{8\sqrt{2}}{3}-\frac{\sqrt{2}}{2}$

Taking LCM $6$: $A=\frac{16\sqrt{2}-3\sqrt{2}}{6}=\frac{13\sqrt{2}}{6}$

Match with options

$\boxed{\frac{13\sqrt{2}}{6}}$

This is Option C.