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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.R.84
Chapter 6, Problem 6.R.84

82–84. Fluid Forces Suppose the following plates are placed on a vertical wall so that the top of the plate is 2 m below the surface of a pool that is filled with water. Compute the force on each plate.


A circular plate with a radius of 2 m

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Identify the depth variable and set up a coordinate system. Let the vertical coordinate \(y\) measure the depth below the water surface, with \(y=0\) at the surface and increasing downward. Since the top of the plate is 2 m below the surface, the top edge corresponds to \(y=2\) m.
Determine the depth range of the plate. The plate is circular with radius 2 m, so its vertical extent is from \(y=2\) m (top) to \(y=2 + 2 \times 2 = 6\) m (bottom), because the diameter is 4 m and the plate extends 2 m above and below its center if centered at 4 m depth. However, since the problem states the top is 2 m below the surface, the plate extends from \(y=2\) m to \(y=6\) m.
Express the width of the plate at a given depth \(y\). For a horizontal slice at depth \(y\), the width of the plate is the length of the chord of the circle at that depth. Using the circle equation centered at \(y=4\) m (midpoint of the plate), the half-width at depth \(y\) is \(\sqrt{2^2 - (y - 4)^2}\). Therefore, the width is \(2 \sqrt{4 - (y - 4)^2}\).
Write the expression for the fluid pressure at depth \(y\). The pressure due to water at depth \(y\) is \(p(y) = \rho g y\), where \(\rho\) is the density of water and \(g\) is the acceleration due to gravity. This pressure acts uniformly across the horizontal slice at depth \(y\).
Set up the integral for the total fluid force on the plate. The force on a thin horizontal strip of thickness \(dy\) at depth \(y\) is \(dF = p(y) \times \text{width}(y) \times dy = \rho g y \times 2 \sqrt{4 - (y - 4)^2} \times dy\). Integrate this expression from \(y=2\) to \(y=6\) to find the total force: \(F = \int_{2}^{6} 2 \rho g y \sqrt{4 - (y - 4)^2} \, dy\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases linearly with depth and is calculated as P = ρgh, where ρ is the fluid density, g is gravitational acceleration, and h is the depth below the surface.

Fluid Force on a Surface

Fluid force on a submerged surface is the total force exerted by the fluid pressure over the area of the surface. It is found by integrating the pressure over the surface area, often simplified by using the average pressure at the centroid multiplied by the area.
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Example 1: Minimizing Surface Area

Geometry of the Plate (Circular Area)

The geometry of the plate determines the area over which the fluid pressure acts. For a circular plate, the area is A = πr², where r is the radius. Knowing the area and depth helps calculate the total fluid force on the plate.
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