The diagram shows the graph of $y = f(x)$. The curve has a turning point at $(0,a)$ and the lines $y = \pm kx$ are asymptotes to the curve, where $a$ and $k$ are positive real constants. [Continued] It is now given that $\mathrm{f}(x) = a\sqrt{1+x^2}$ and this holds for the rest of the question.

1. Deduce the relationship between the constants $a$ and $k$.

Given $\mathrm{f}(x) = a\sqrt{1+x^2}$, for large values of $|x|$, $\mathrm{f}(x) \approx a\sqrt{x^2} = a|x|$.
Thus, the asymptotes of $y = \mathrm{f}(x)$ are $y = ax$ and $y = -ax$.
The question states the asymptotes are $y = \pm kx$, where $a$ and $k$ are positive real constants. By comparing the equations, we deduce that $k = a$.

2. Determine the key features of the graph of $y = \frac{1}{k}\mathrm{f}(x+2a)$.

Substitute $k=a$: $y = \frac{1}{a}\mathrm{f}(x+2a) = \sqrt{1+(x+2a)^2}$.
This corresponds to translating the graph of $\mathrm{f}(x)$ by $2a$ units in the negative $x$-direction, followed by a vertical scaling by $\frac{1}{a}$.

• Turning point: The original turning point $(0,a)$ becomes $(-2a, a)$, then scales to $(-2a, 1)$.
• Asymptotes: The original asymptotes $y = \pm a(x)$ translate to $y = \pm a(x+2a)$ and scale by $\frac{1}{a}$ to become $y = x+2a$ and $y = -x-2a$.
• $y$-intercept: Substitute $x=0$, $y = \sqrt{1+(2a)^2} = \sqrt{1+4a^2}$.

3. Differentiate $\mathrm{f}(x)$ to find the expression for the second graph.

$\mathrm{f}(x) = a(1+x^2)^{\frac{1}{2}}$
Using the chain rule:
$\mathrm{f}'(x) = a \left( \frac{1}{2} \right)(1+x^2)^{-\frac{1}{2}}(2x) = \frac{ax}{\sqrt{1+x^2}}$

4. Determine the key features of the graph of $y = \mathrm{f}'(x)$.

• Intersection with axes: When $x=0$, $y=0$, so the curve passes through $(0,0)$.
• Asymptotes:
  As $x \to \infty$, $y = \frac{ax}{x\sqrt{\frac{1}{x^2}+1}} \to a$.
  As $x \to -\infty$, $y = \frac{ax}{-x\sqrt{\frac{1}{x^2}+1}} \to -a$.
  Hence, horizontal asymptotes are $y = a$ and $y = -a$.
• Turning points: Since the curve strictly transitions from asymptotic to $-a$ up to $a$, and $\mathrm{f}''(x) = \frac{a}{(1+x^2)^{3/2}} > 0$ for all $x$, there are no turning points.

1. Determine the boundary $y$-values for the region by evaluating the function at the $x$-limits.

The region is bounded by $x = 0$ (the $y$-axis) and $x = \sqrt{3}$. Find the corresponding $y$-values on the curve $y = a\sqrt{1+x^2}$:
When $x = 0$, $y = a\sqrt{1+0^2} = a$.
When $x = \sqrt{3}$, $y = a\sqrt{1+(\sqrt{3})^2} = a\sqrt{4} = 2a$.

2. Express $x^2$ in terms of $y$ to prepare for volume integration with respect to $y$.

$y = a\sqrt{1+x^2}$
$y^2 = a^2(1+x^2)$
$\frac{y^2}{a^2} = 1 + x^2$
$x^2 = \frac{y^2}{a^2} - 1$

3. Set up the integrals for the volume of revolution using the disk and washer methods.

The region $R$ rotated about the $y$-axis can be split into two sections vertically:

1. For $0 \le y \le a$, the cross-sections are solid disks of radius $R = \sqrt{3}$.
2. For $a \le y \le 2a$, the cross-sections are washers with an outer radius $R_{out} = \sqrt{3}$ and an inner radius $R_{in} = x$.

$V = \int_{0}^{a} \pi (\sqrt{3})^2 \, dy + \int_{a}^{2a} \pi \left( (\sqrt{3})^2 - x^2 \right) \, dy$
$V = [3\pi y]_0^a + \pi \int_{a}^{2a} \left( 3 - \left(\frac{y^2}{a^2} - 1\right) \right) \, dy$
$V = 3\pi a + \pi \int_{a}^{2a} \left( 4 - \frac{y^2}{a^2} \right) \, dy$

4. Evaluate the definite integral to find the exact final volume.

$\int_{a}^{2a} \left( 4 - \frac{y^2}{a^2} \right) \, dy = \left[ 4y - \frac{y^3}{3a^2} \right]_a^{2a}$
$= \left( 4(2a) - \frac{(2a)^3}{3a^2} \right) - \left( 4(a) - \frac{a^3}{3a^2} \right)$
$= \left( 8a - \frac{8a}{3} \right) - \left( 4a - \frac{a}{3} \right)$
$= \frac{16a}{3} - \frac{11a}{3} = \frac{5a}{3}$

Therefore, the total volume of the sculpture is:
$V = 3\pi a + \pi \left(\frac{5a}{3}\right) = \frac{14}{3}\pi a$