Hydrostatic Pressure Load on a Parabolic Dam Floodgate (Civil & Environmental Engineering) Solution

A water resources and structural civil engineer at the UNLV Howard R. Hughes College of Engineering is analyzing the structural integrity of a submerged irrigation canal floodgate near Lake Mead.

The vertical floodgate is modeled as a planar general region $R$ in the $xy$-plane bounded on the left by the parabolic retaining edge $x = y^2$ and on the right by the vertical concrete support wall $x = 4$, with spatial dimensions measured in meters ($0 \le y \le 2$):

The water surface reaches the crest of the floodgate at $y = 2\text{ m}$. Hydrostatic fluid pressure $P(x, y)$ increases linearly with submersion depth $(2 - y)$ according to the hydrostatic law:

where $\rho = 1000\text{ kg/m}^3$ is fresh water density and $g = 9.81\text{ m/s}^2$ is gravitational acceleration.

Step-by-Step Solution:

1. Classify the Planar Region: State the Type II General Region Definition (defined in Double Integrals over a General Region): Because the left boundary is defined explicitly as a function of $y$ ($x = y^2$) and the right boundary is the constant line $x = 4$, slicing horizontally across constant $y$-levels ($dx \, dy$) eliminates the need to invert functions. The region is naturally of Type II:

   $$R = \{(x, y) \in \mathbb{R}^2 \mid 0 \le y \le 2, \, y^2 \le x \le 4\}$$

2. Formulate the Resultant Hydrostatic Force Integral: State the Hydrostatic Pressure Force Integral Formula:

   $$F_R = \iint_R P(x, y) \, dA = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} \rho g (2 - y) \, dx \, dy$$
   Substitute the Type II limits:
   $$F_R = \rho g \int_{0}^{2} \int_{y^2}^{4} (2 - y) \, dx \, dy$$

3. Evaluate the Inner Integral ($dx$): Holding $y$ constant, $(2 - y)$ acts as a scalar multiplier:

   $$\int_{y^2}^{4} (2 - y) \, dx = (2 - y) [x]_{y^2}^{4} = (2 - y)(4 - y^2)$$
   Expand the polynomial product:
   $$(2 - y)(4 - y^2) = 8 - 2y^2 - 4y + y^3 = 8 - 4y - 2y^2 + y^3$$

4. Evaluate the Outer Integral ($dy$):

   $$F_R = \rho g \int_{0}^{2} (8 - 4y - 2y^2 + y^3) \, dy$$
   Find the antiderivative:
   $$F_R = \rho g \left[ 8y - 2y^2 - \frac{2}{3}y^3 + \frac{1}{4}y^4 \right]_{0}^{2}$$
   Evaluate at the upper limit $y = 2$:
   $$F_R = \rho g \left( 8(2) - 2(2)^2 - \frac{2}{3}(2)^3 + \frac{1}{4}(2)^4 \right)$$
   $$= \rho g \left( 16 - 8 - \frac{16}{3} + 4 \right) = \rho g \left( 12 - \frac{16}{3} \right)$$
   $$= \rho g \left( \frac{36}{3} - \frac{16}{3} \right) = \frac{20}{3} \rho g\text{ N}$$

5. Compute Numerical Resultant Load: Substitute $\rho = 1000\text{ kg/m}^3$ and $g = 9.81\text{ m/s}^2$:

   $$F_R = \frac{20}{3}(1000)(9.81) = \frac{196,200}{3} = 65,400\text{ N} = 65.4\text{ kN}$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The total horizontal hydrostatic force exerted by the impounded water against the curved parabolic gate is $F_R = \frac{20}{3}\rho g \approx 65.4\text{ kN}$.
• Engineering Application: In civil hydraulic infrastructure, water pressure intensifies with depth, creating an asymmetrical triangular loading prism. Slicing horizontally as a Type II double integral allows hydraulic engineers to easily determine both the magnitude and the vertical line of action (center of pressure) of the resultant force, ensuring that emergency spillway gate trunnion pins and hydraulic lift cylinders are sized with adequate structural safety factors against shear failure.