喷嘴通常用于软管的末端，以引导和加速水流。下图表示一个喷嘴的侧视图。 喷嘴本身由外表面和内表面之间的固体材料组成（图中阴影部分）。水流过由内表面包围的空心区域 ABCD。喷嘴的外表面是通过将方程为 $y^{2} = k + \dfrac{2k}{(x+1)(x+2)},\qquad\text{for }0\le x\le k,\text{ where }k\in\mathbb{R},\;k\ge 1,$ 的曲线绕 x 轴完整旋转一周而得到的。 类似地，内表面是通过将方程为 $y^{2} = k - \dfrac{x}{3},\qquad\text{for }0\le x\le k,$ 的曲线绕 x 轴完整旋转一周而得到的。$x$ 和 $y$ 的单位是厘米。 利用质量守恒定律，可以得到以下关系式。 $A_{\text{in}}v_{\text{in}} = A_{\text{out}}v_{\text{out}}$ 其中 $A_{\text{in}}$ 和 $A_{\text{out}}$ 分别是喷嘴入口和出口处圆形开口的横截面积，且 $v_{\text{in}}$ 和 $v_{\text{out}}$ 分别是喷嘴入口和出口处的水流速度。

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:
$\frac{v_{\text{out}}}{v_{\text{in}}} = \frac{\pi k}{\frac{2\pi k}{3}}$
$\frac{v_{\text{out}}}{v_{\text{in}}} = \frac{3}{2}$