1. Given curves and lines

We have:

• Parabola: $y^2=8x$
• Line: $y=x$
• Vertical line: $x=2$

We want the larger bounded region in the first quadrant.

2. Find intersection points

(i) Intersection of $y=x$ and $y^2=8x$

Substitute $y=x$ into $y^2=8x$:

$x^2=8x$ $x(x-8)=0$

So, $x=0$ or $x=8$. Since $y=x$, intersection points are:

$(0,0), \quad (8,8)$

(ii) Intersection of $x=2$ with the parabola

From $y^2=8x$:

$y^2=16 \Rightarrow y=4$

(first quadrant), so point is $(2,4)$.

(iii) Intersection of $x=2$ with $y=x$

This gives point $(2,2)$.

3. Identify the bounded region in the first quadrant

Between $x=2$ and $x=8$, the line $y=x$ and the upper branch of parabola

$y=\sqrt{8x}$

form a closed region together with the line $x=2$.

At $x=2$:

• line gives $y=2$
• parabola gives $y=4$

So the enclosed region has vertical slice height:

$\text{upper} - \text{lower} = \sqrt{8x}-x$

for $2 \le x \le 8$.

This is the larger bounded region.

4. Compute the area $\alpha$

$\alpha=\int_2^8 (\sqrt{8x}-x)\,dx$

Now,

$\sqrt{8x}=2\sqrt{2}\,x^{1/2}$

Hence,

$\alpha=\int_2^8 2\sqrt{2}x^{1/2}\,dx - \int_2^8 x\,dx$

Compute each integral:

$\int 2\sqrt{2}x^{1/2}\,dx = 2\sqrt{2}\cdot \frac{2}{3}x^{3/2}=\frac{4\sqrt{2}}{3}x^{3/2}$

$\int x\,dx=\frac{x^2}{2}$

So,

$\alpha=\left[\frac{4\sqrt{2}}{3}x^{3/2}-\frac{x^2}{2}\right]_2^8$

Now evaluate:

For $x=8$:

$8^{3/2}=8\sqrt{8}=8\cdot 2\sqrt{2}=16\sqrt{2}$

Thus,

$\frac{4\sqrt{2}}{3}\cdot 16\sqrt{2}=\frac{4\cdot 16\cdot 2}{3}=\frac{128}{3}$

and

$\frac{8^2}{2}=32$

So value at $x=8$ is

$\frac{128}{3}-32=\frac{128-96}{3}=\frac{32}{3}$

For $x=2$:

$2^{3/2}=2\sqrt{2}$

Thus,

$\frac{4\sqrt{2}}{3}\cdot 2\sqrt{2}=\frac{4\cdot 2\cdot 2}{3}=\frac{16}{3}$

and

$\frac{2^2}{2}=2$

So value at $x=2$ is

$\frac{16}{3}-2=\frac{16-6}{3}=\frac{10}{3}$

Therefore,

$\alpha=\frac{32}{3}-\frac{10}{3}=\frac{22}{3}$

5. Find $3\alpha$

$3\alpha=3\cdot \frac{22}{3}=22$

6. Comparison with stored answer

Stored correct answer = $22$.

Our derived answer is also $22$, so it agrees.