Understanding Ocean Water Pressure Calculation

The problem asks us to find the formula for pressure ($P$) as a function of depth ($h$) in ocean water. We are given how the specific weight ($W$) of the water changes with depth: $$W = W_0 + C\sqrt{h}$$ Here, $W_0$ represents the specific weight at the surface, and $C$ is a constant value.

Deriving the Pressure Formula

In fluid mechanics, pressure ($P$) at a certain depth is calculated by integrating the specific weight ($W$) over the depth ($h$). The relationship is given by the integral:

$$P = \int W \, dh$$

Now, we substitute the given expression for the specific weight ($W$) into this integral:

$$P = \int (W_0 + C\sqrt{h}) \, dh$$

We can split this integral into two parts:

$$P = \int W_0 \, dh + \int C\sqrt{h} \, dh$$

Let's solve each part:

• For the first part, integrating $W_0$ with respect to $h$ (since $W_0$ is a constant): $$\int W_0 \, dh = W_0h$$
• For the second part, integrating $C\sqrt{h}$ with respect to $h$. First, we can rewrite $\sqrt{h}$ as $h^{1/2}$: $$\int C h^{1/2} \, dh$$ Since $C$ is a constant, we can take it out of the integral: $$C \int h^{1/2} \, dh$$ Using the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1}$, where $n = 1/2$: $$C \frac{h^{(1/2) + 1}}{(1/2) + 1} = C \frac{h^{3/2}}{3/2}$$ Simplifying the fraction $\frac{1}{3/2}$: $$C \times \frac{2}{3} h^{3/2} = \frac{2}{3} C h^{3/2}$$

Final Pressure Equation

Now, we combine the results from both parts of the integration to get the final expression for pressure ($P$) as a function of depth ($h$):

$$P = W_0h + \frac{2}{3} C h^{3/2}$$

This equation represents the total pressure at depth $h$, considering the variation in specific weight.