Nozzles are often used at the end of hoses to direct and speed up the flow of water. The diagram below represents the side view of a nozzle. The nozzle itself consists of solid material between an outer surface and an inner surface (shaded in the diagram). Water flows through the hollow region ABCD enclosed by the inner surface. The outer surface of the nozzle is obtained by rotating the curve with equation $y^{2} = k + \dfrac{2k}{(x+1)(x+2)},\qquad\text{for }0\le x\le k,\text{ where }k\in\mathbb{R},\;k\ge 1,$ completely about the x-axis. Similarly, the inner surface is obtained by rotating the curve with equation $y^{2} = k - \dfrac{x}{3},\qquad\text{for }0\le x\le k,$ completely about the x-axis. The units of $x$ and $y$ are centimetres. Using the Principle of Conservation of Mass, the following relationship can be found. $A_{\text{in}}v_{\text{in}} = A_{\text{out}}v_{\text{out}}$ $A_{\text{in}}$ and $A_{\text{out}}$ are the cross-sectional areas of the circular openings at the inlet and outlet of the nozzle respectively, and $v_{\text{in}}$ and $v_{\text{out}}$ are the speeds of the water at the inlet and outlet of the nozzle respectively.

1. Set up the definite integral for the volume of the solid of revolution, which is the difference between the outer and inner volumes.

$V = \pi \int_{0}^{k} y_{\text{outer}}^2 \, dx - \pi \int_{0}^{k} y_{\text{inner}}^2 \, dx$
$V = \pi \int_{0}^{k} \left[ \left( k + \frac{2k}{(x+1)(x+2)} \right) - \left( k - \frac{x}{3} \right) \right] \, dx$
$V = \pi \int_{0}^{k} \left[ \frac{2k}{(x+1)(x+2)} + \frac{x}{3} \right] \, dx$

2. Decompose the rational expression into partial fractions to prepare for integration.

$\frac{2k}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$
$2k = A(x+2) + B(x+1)$
Let $x = -1$: $A = 2k$
Let $x = -2$: $B = -2k$
So, $\frac{2k}{(x+1)(x+2)} = \frac{2k}{x+1} - \frac{2k}{x+2}$

3. Find the antiderivative of the simplified expression.

$V = \pi \int_{0}^{k} \left( \frac{2k}{x+1} - \frac{2k}{x+2} + \frac{x}{3} \right) \, dx$
$V = \pi \left[ 2k \ln|x+1| - 2k \ln|x+2| + \frac{x^2}{6} \right]_{0}^{k}$

4. Evaluate the definite integral by substituting the upper and lower limits.

Upper limit ($x = k$):
$2k \ln(k+1) - 2k \ln(k+2) + \frac{k^2}{6}$
Lower limit ($x = 0$):
$2k \ln(1) - 2k \ln(2) + 0 = -2k \ln 2$
Subtracting the lower limit from the upper limit:
$V = \pi \left( 2k \ln(k+1) - 2k \ln(k+2) + \frac{k^2}{6} - (-2k \ln 2) \right)$

5. Use laws of logarithms to combine the terms and reach the target expression.

$V = \pi \left( 2k \left[ \ln(k+1) - \ln(k+2) + \ln 2 \right] + \frac{k^2}{6} \right)$
$V = \pi \left( 2k \ln\left( \frac{2(k+1)}{k+2} \right) + \frac{k^2}{6} \right)$
$V = \pi \left[ 2k \ln\left( \frac{2k+2}{k+2} \right) + \frac{k^2}{6} \right] \text{ cm}^3 \text{ (shown)}$

1. Set up the integral for the area of the 2D hollow region $ABCD$.

From the side view, the hollow region $ABCD$ is enclosed by the inner surface. This means it is bounded symmetrically above and below the $x$-axis by the curves $y = \sqrt{k - \frac{x}{3}}$ and $y = -\sqrt{k - \frac{x}{3}}$ for $0 \le x \le k$.

$\text{Area} = \int_{0}^{k} \left[ \sqrt{k - \frac{x}{3}} - \left(-\sqrt{k - \frac{x}{3}}\right) \right] \, \mathrm{d}x = 2 \int_{0}^{k} \left( k - \frac{x}{3} \right)^{\frac{1}{2}} \, \mathrm{d}x$

2. Integrate the function using the standard rule for fractional powers of a linear function.

$\int \left( k - \frac{x}{3} \right)^{\frac{1}{2}} \, \mathrm{d}x = \frac{\left( k - \frac{x}{3} \right)^{\frac{3}{2}}}{\frac{3}{2} \cdot \left(-\frac{1}{3}\right)} = -2 \left( k - \frac{x}{3} \right)^{\frac{3}{2}}$

Therefore, the area formula becomes:
$\text{Area} = 2 \left[ -2 \left( k - \frac{x}{3} \right)^{\frac{3}{2}} \right]_{0}^{k} = -4 \left[ \left( k - \frac{x}{3} \right)^{\frac{3}{2}} \right]_{0}^{k}$

3. Evaluate the definite integral by substituting the upper and lower limits.

Substitute the upper limit $x = k$:
$-4 \left( k - \frac{k}{3} \right)^{\frac{3}{2}} = -4 \left( \frac{2k}{3} \right)^{\frac{3}{2}}$

Substitute the lower limit $x = 0$:
$-4 \left( k - 0 \right)^{\frac{3}{2}} = -4k^{\frac{3}{2}}$

Subtract the lower limit evaluation from the upper limit:
$\text{Area} = -4 \left( \frac{2k}{3} \right)^{\frac{3}{2}} - \left( -4k^{\frac{3}{2}} \right) = 4k^{\frac{3}{2}} - 4 \left( \frac{2}{3} \right)^{\frac{3}{2}} k^{\frac{3}{2}}$

Factor and simplify the constants:
$\text{Area} = 4k^{\frac{3}{2}} \left( 1 - \frac{2\sqrt{2}}{3\sqrt{3}} \right) = 4k^{\frac{3}{2}} \left( 1 - \frac{2\sqrt{6}}{9} \right)$

1. Relate the ratio of speeds to the ratio of cross-sectional areas and find the area at the inlet.

From the Principle of Conservation of Mass, $A_{\text{in}}v_{\text{in}} = A_{\text{out}}v_{\text{out}}$, so:
$\frac{v_{\text{out}}}{v_{\text{in}}} = \frac{A_{\text{in}}}{A_{\text{out}}}$
The water flows through the hollow region enclosed by the inner surface, so the cross-sections are circular with area $A = \pi y^2$, where $y^2 = k - \frac{x}{3}$.

At the inlet ($x = 0$):
$A_{\text{in}} = \pi \left( k - \frac{0}{3} \right) = \pi k$

2. Find the cross-sectional area at the outlet.

At the outlet ($x = k$):
$A_{\text{out}} = \pi \left( k - \frac{k}{3} \right) = \frac{2\pi k}{3}$

3. Calculate the ratio of the speeds.

Substituting the areas into the ratio equation:
$\frac{v_{\text{out}}}{v_{\text{in}}} = \frac{\pi k}{\frac{2\pi k}{3}}$
$\frac{v_{\text{out}}}{v_{\text{in}}} = \frac{3}{2}$