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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 6.7.54
Chapter 6, Problem 6.7.54

52–54. Force on a window A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows. 


The window is circular, with a radius of 0.5 m, tangent to the bottom of the pool.

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Identify the physical context: The force on the window is due to the water pressure acting on its surface. Water pressure increases with depth according to the formula \(P = \rho g h\), where \(\rho\) is the density of water, \(g\) is the acceleration due to gravity, and \(h\) is the depth below the water surface.
Set up a coordinate system: Let the vertical coordinate \(y\) measure the depth from the water surface downward. Since the pool is 4 m deep, the bottom is at \(y = 4\). The circular window is tangent to the bottom, so its lowest point is at \(y = 4\) and its radius is 0.5 m.
Express the shape of the window: The window is a circle of radius 0.5 m tangent to the bottom at \(y=4\). The vertical extent of the window goes from \(y=3.5\) (top of the circle) to \(y=4\) (bottom). For each depth \(y\) in this interval, find the horizontal width of the window slice, which corresponds to the length of a horizontal chord of the circle at depth \(y\).
Calculate the pressure at depth \(y\): Use \(P(y) = \rho g y\), where \(y\) is the depth from the surface. This pressure acts uniformly across the horizontal slice of the window at depth \(y\).
Set up the integral for the total force: The force on a thin horizontal strip of the window at depth \(y\) is \(dF = P(y) \times \text{width}(y) \times dy\). Integrate this expression from \(y=3.5\) to \(y=4\) to find the total force on the window.

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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.

Force on a Submerged Surface

The force on a submerged surface is found by integrating the pressure over the area of the surface. Since pressure varies with depth, the total force equals the pressure at the centroid of the area times the area, accounting for the fluid’s depth distribution.
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Example 1: Minimizing Surface Area

Geometry of the Circular Window

Understanding the position and size of the circular window is essential to determine the depth of its centroid and the limits of integration. Since the window is tangent to the bottom, its center is located at a radius above the bottom, affecting the pressure distribution.
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Example 4: Norman Window
Related Practice
Textbook Question

Why is integration used to find the work required to pump water out of a tank?

Textbook Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

x = 4 / y + y³,x = 1/√3, and y=1; about the x-axis

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Textbook Question

Find the area of the region described in the following exercises.


The region in the first quadrant bounded by y=x^2/3 and y=4

Textbook Question

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.


x =2

Textbook Question

13–20. Mass of one-dimensional objects Find the mass of the following thin bars with the given density function.


ρ(x) = {1 if 0≤x≤2 {2 if 2<x≤3

Textbook Question

Suppose f and g have continuous derivatives on an interval [a, b]. Prove that if f(a)=g(a) and f(b)=g(b), then ∫a^b f′(x) dx = ∫a^b g′(x) dx.