Given Information:

• The triangular plate is submerged 22 ft below the surface of a freshwater lake.
• The triangle is isosceles with a base of 6 ft and a height of 6 ft.
• Freshwater has a weight density of $\rho = 62.4 \, \text{lb/ft}^3$.
• We are asked to find the fluid force against one face of the plate using integrals.

Step 1: Coordinate System and Geometry Setup

To simplify the problem, we place the origin of the coordinate system at the bottom vertex of the triangle. The $y$-axis runs upward along the plate's axis of symmetry.

• The height of the plate (from the bottom vertex to the base) is 6 ft.
• The base of the plate is 6 ft wide, which means the half-width of the plate at any height $y$ is 3 ft (symmetrical about the $y$-axis).

The equation for the half-width of the plate at height $y$ can be derived using the geometry of the triangle. Since the plate is isosceles: $\frac{x(y)}{6} = \frac{y}{6} \quad \text{(linear relationship between width and height)}$ This simplifies to: $x(y) = y$ So, the total width of the plate at height $y$ is: $\text{Width at } y = 2x(y) = 2y$

Step 2: Depth of the Plate

The plate is submerged