Submersible Pressure Hull Buoyancy and Hydrostatic Stability (Mechanical & Ocean Engineering) Solution

A mechanical and ocean systems engineer at the UNLV Howard R. Hughes College of Engineering is modeling the hydrodynamic stability of an unmanned underwater vehicle (UUV) designed for deep-sea trench exploration.

The symmetric profile of the submerged starboard hull section is bounded on the left by the vertical keel centerline $x = 0$, at the top by the design waterline $y = 3$, and on the exterior by the logarithmic hull boundary $x = \ln(y)$ (for $y \ge 1$, dimensions in meters). The static hydrostatic buoyant pressure moment is initially modeled as a Type II horizontal integral:

Step-by-Step Solution:

1. Extract Original Boundaries and Invert Boundary Function: State the original Type II Boundaries:

   $$1 \le y \le 3, \quad 0 \le x \le \ln(y)$$
   Invert the boundary function $x = \ln(y)$ to express $y$ as a function of $x$:
   $$x = \ln(y) \iff y = e^x$$

2. Re-slice as a Type I General Region: Apply the 4-Step Re-slicing Protocol (defined in Switching the Order of Integration):

   • Determine the horizontal extent of the domain: at $y = 1 \implies x = \ln(1) = 0$; at $y = 3 \implies x = \ln(3)$.
   • The outer constant interval for $x$ is $0 \le x \le \ln(3)$.
   • For any fixed $x$, a vertical slice enters at the lower hull curve $y = e^x$ and exits at the horizontal waterline $y = 3$. Thus, the Type I formulation is:
     $$R = \{(x, y) \in \mathbb{R}^2 \mid 0 \le x \le \ln(3), \, e^x \le y \le 3\}$$

3. Formulate and Evaluate Switched Double Integral: State the Type I Iterated Integral:

   $$M = \int_{0}^{\ln(3)} \int_{e^x}^{3} y \, dy \, dx$$

   Evaluate Inner Integral ($dy$):

   $$\int_{e^x}^{3} y \, dy = \left[ \frac{1}{2}y^2 \right]_{y=e^x}^{y=3} = \frac{1}{2}(3)^2 - \frac{1}{2}(e^x)^2 = \frac{9}{2} - \frac{1}{2}e^{2x}$$

   Evaluate Outer Integral ($dx$):

   $$M = \int_{0}^{\ln(3)} \left( \frac{9}{2} - \frac{1}{2}e^{2x} \right) \, dx = \left[ \frac{9}{2}x - \frac{1}{4}e^{2x} \right]_{0}^{\ln(3)}$$
   Evaluate at the boundary limits:
   $$M = \left( \frac{9}{2}\ln(3) - \frac{1}{4}e^{2\ln(3)} \right) - \left( \frac{9}{2}(0) - \frac{1}{4}e^0 \right)$$
   Simplify exponential terms using $e^{2\ln(3)} = e^{\ln(3^2)} = 3^2 = 9$ and $e^0 = 1$:
   $$M = \left( \frac{9}{2}\ln(3) - \frac{9}{4} \right) - \left( 0 - \frac{1}{4} \right) = \frac{9}{2}\ln(3) - \frac{9}{4} + \frac{1}{4}$$
   $$M = \frac{9}{2}\ln(3) - \frac{8}{4} = \frac{9}{2}\ln(3) - 2$$
   Approximate numerically with $\ln(3) \approx 1.098612$:
   $$M = \frac{9}{2}(1.098612) - 2 = 4.943755 - 2 \approx 2.944\text{ m}^3$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The static hydrostatic buoyant pressure moment of the submerged starboard hull profile is $M = \frac{9}{2}\ln(3) - 2 \approx 2.944\text{ m}^3$.
• Engineering Application: When an underwater autonomous exploration vehicle tilts due to ocean currents or ballast shifts, the center of buoyancy $\overline{KB}$ shifts laterally. Ocean and mechanical engineers calculate area moments across switched integration frames to dynamically evaluate the vessel's transverse metacentric height $\overline{GM} = \overline{KB} + \overline{BM} - \overline{KG}$, ensuring that the submersible possesses an adequate restoring righting moment to resist roll instability and prevent capsizing in turbulent deep-sea currents.