1. Interpret the region

We are given $S=\{(x,y): y^2\le 8x,\; y\ge \sqrt{2}\,x,\; x\ge 1\}.$

The curves/boundaries are:

• Parabola: $y^2=8x \implies x=\frac{y^2}{8}$
• Line: $y=\sqrt{2}\,x \implies x=\frac{y}{\sqrt{2}}$
• Vertical line: $x=1$

Also, since $y\ge \sqrt{2}x$ and $x\ge 1$, we have $y\ge \sqrt{2}$.

2. Understand the inequalities

From $y^2\le 8x,$ we get $x\ge \frac{y^2}{8}.$

From $y\ge \sqrt{2}x,$ we get $x\le \frac{y}{\sqrt{2}}.$

And also $x\ge 1.$

So for a fixed $y$, the allowed $x$ satisfy $\max\left(1,\frac{y^2}{8}\right) \le x \le \frac{y}{\sqrt{2}}.$

Hence we must determine the $y$-range where this is possible.

3. Find relevant intersection points

(i) Line and parabola

Solve $y=\sqrt{2}x,\qquad y^2=8x.$ Substitute $x=\frac{y}{\sqrt{2}}$ into $y^2=8x$: $y^2=8\cdot \frac{y}{\sqrt{2}}=4\sqrt{2}\,y.$ So, $y(y-4\sqrt{2})=0.$ Thus the positive intersection is $y=4\sqrt{2},\qquad x=\frac{y}{\sqrt{2}}=4.$

(ii) Line and $x=1$

At $x=1$, $y=\sqrt{2}.$ So point is $(1,\sqrt{2})$.

(iii) Parabola and $x=1$

At $x=1$, $y^2=8 \implies y=2\sqrt{2} \quad (\text{upper branch relevant}).$ So point is $(1,2\sqrt{2})$.

4. Describe the region by horizontal strips

For $y$ between $\sqrt{2}$ and $2\sqrt{2}$:

• $\frac{y^2}{8}\le 1,$ so left boundary is $x=1$.
• Right boundary is $x=\frac{y}{\sqrt{2}}$.

For $y$ between $2\sqrt{2}$ and $4\sqrt{2}$:

• $\frac{y^2}{8}\ge 1,$ so left boundary is $x=\frac{y^2}{8}$.
• Right boundary is $x=\frac{y}{\sqrt{2}}$.

Therefore, $\text{Area} = \int_{\sqrt{2}}^{2\sqrt{2}}\left(\frac{y}{\sqrt{2}}-1\right)dy + \int_{2\sqrt{2}}^{4\sqrt{2}}\left(\frac{y}{\sqrt{2}}-\frac{y^2}{8}\right)dy.$

5. Compute the first integral

$I_1=\int_{\sqrt{2}}^{2\sqrt{2}}\left(\frac{y}{\sqrt{2}}-1\right)dy.$

Antiderivative: $\int \left(\frac{y}{\sqrt{2}}-1\right)dy=\frac{y^2}{2\sqrt{2}}-y.$

So, $I_1=\left[\frac{y^2}{2\sqrt{2}}-y\right]_{\sqrt{2}}^{2\sqrt{2}}.$

At $y=2\sqrt{2}$: $\frac{(2\sqrt{2})^2}{2\sqrt{2}}-2\sqrt{2}=\frac{8}{2\sqrt{2}}-2\sqrt{2}=\frac{4}{\sqrt{2}}-2\sqrt{2}=2\sqrt{2}-2\sqrt{2}=0.$

At $y=\sqrt{2}$: $\frac{(\sqrt{2})^2}{2\sqrt{2}}-\sqrt{2}=\frac{2}{2\sqrt{2}}-\sqrt{2}=\frac{1}{\sqrt{2}}-\sqrt{2}=-\frac{\sqrt{2}}{2}.$

Thus, $I_1=0-\left(-\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}.$

6. Compute the second integral

$I_2=\int_{2\sqrt{2}}^{4\sqrt{2}}\left(\frac{y}{\sqrt{2}}-\frac{y^2}{8}\right)dy.$

Antiderivative: $\int \left(\frac{y}{\sqrt{2}}-\frac{y^2}{8}\right)dy=\frac{y^2}{2\sqrt{2}}-\frac{y^3}{24}.$

So, $I_2=\left[\frac{y^2}{2\sqrt{2}}-\frac{y^3}{24}\right]_{2\sqrt{2}}^{4\sqrt{2}}.$

At $y=4\sqrt{2}$: \frac{(4\sqrt{2})^2}{2\sqrt{2}}-\frac{(4\sqrt{2})^3}{24}=rac{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}.

At $y=2\sqrt{2}$: \frac{(2\sqrt{2})^2}{2\sqrt{2}}-\frac{(2\sqrt{2})^3}{24}=rac{8}{2\sqrt{2}}-\frac{16\sqrt{2}}{24}=2\sqrt{2}-\frac{2\sqrt{2}}{3}=\frac{4\sqrt{2}}{3}.

Thus, $I_2=\frac{8\sqrt{2}}{3}-\frac{4\sqrt{2}}{3}=\frac{4\sqrt{2}}{3}.$

7. Total area

$\text{Area}=I_1+I_2=\frac{\sqrt{2}}{2}+\frac{4\sqrt{2}}{3}.$

Taking LCM $6$: $\text{Area}=\frac{3\sqrt{2}}{6}+\frac{8\sqrt{2}}{6}=\frac{11\sqrt{2}}{6}.$

8. Compare with options

The correct option is $\boxed{\frac{11\sqrt{2}}{6}}$ which is Option B.

9. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.