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HomeCalculus II: Integral Calculus - Integration, Series, and Parametric EquationsChapter Study Guide: Physical Applications of Integration
Study Guide925 words

Chapter Study Guide: Physical Applications of Integration

Physical Applications

Learning Objectives

After completing this section, you should be able to:

  • Determine the mass of a one-dimensional object from its linear density function.
  • Determine the mass of a two-dimensional circular object from its radial density function.
  • Calculate the work done by a variable force acting along a line.
  • Calculate the work done in pumping a liquid from one height to another.
  • Find the hydrostatic force against a submerged vertical plate.

Key Terms & Glossary

  • **Linear Density Function ρ(x):∗∗Anintegrablefunctiondescribingthemassperunitlengthofanobjectatanyspecificpointx\rho(x):** An integrable function describing the mass per unit length of an object at any specific point xρ(x):∗∗Anintegrablefunctiondescribingthemassperunitlengthofanobjectatanyspecificpointx.
  • Hydrostatic Force: The force exerted by a stationary liquid on a submerged surface, driven by gravity and depth-dependent pressure.
  • Pascal's Principle: A fluid mechanics principle stating that pressure at a given depth is the same in all directions.
  • Riemann Sum: An approximation of the total value of a varying quantity (like mass or force) by partitioning the object into small, constant-value segments.
  • Center of Mass: The theoretical balancing point of a system where total mass could be concentrated without changing the system's moment.

The "Big Idea"

The fundamental "Big Idea" of this chapter is that integration is the ultimate tool for accumulating continuously varying physical quantities. Whether an object's density changes from one end to the other, a force fluctuates as an object is moved, or water pressure increases as you go deeper underwater, the solution strategy is identical. We partition the object or path into infinitesimally small segments (Δx)\Delta x)Δx), assume the quantity is constant over that tiny segment to find a local value, sum those values using a Riemann sum, and takethelimitasΔx→0take the limit as \Delta x \to 0takethelimitasΔx→0 to produce a definite integral.

Formula / Concept Box

Application AreaIntegral SetupKey Components
Mass (1D Object)m=∫abρ(x)dxm = \int_a^b \rho(x) dxm=∫ab​ρ(x)dxρ(x)=lineardensityfunction<br>[a,b]\rho(x) = linear density function <br> [a, b]ρ(x)=lineardensityfunction<br>[a,b] = span of the object
Work (Variable Force)W=∫abF(x)dxW = \int_a^b F(x) dxW=∫ab​F(x)dxF(x)=forceappliedatpositionxF(x) = force applied at position xF(x)=forceappliedatpositionx
[a,b][a, b][a,b] = distance moved
Hydrostatic ForceF=∫abρg⋅s(x)⋅w(x)dxF = \int_a^b \rho g \cdot s(x) \cdot w(x) dxF=∫ab​ρg⋅s(x)⋅w(x)dxρg=weightdensityoffluid<br>s(x)\rho g = weight density of fluid <br> s(x)ρg=weightdensityoffluid<br>s(x) = depth, w(x)w(x)w(x) = width
Moment (1D System)M0=∑i=1nmixiM_0 = \sum_{i=1}^n m_i x_iM0​=∑i=1n​mi​xi​mi=massofparticleim_i = mass of particle imi​=massofparticlei
xix_ixi​ = position from origin

Hierarchical Outline

  • Mass and Density
    • Constant Density: Evaluated simply by multiplication (m=ρLm = \rho Lm=ρL).
    • Variable Linear Density: Requires integrating the linear density function ρ(x)\rho(x)ρ(x) over the length of the rod.
    • Radial Density: Applied to 2D circular objects where density varies outward from the center.
  • Work Applications
    • Variable Force: Integrating a fluctuating force function over a distance (e.g., stretching springs).
    • Pumping Liquids: Calculating work by integrating the weight of fluid layers moved over specific vertical distances.
  • Hydrostatic Force
    • Depth and Pressure: By Pascal's principle, pressure depends strictly on depth.
    • Vertical Submersion: Depth varies along a vertical plate, requiring integration of thin horizontal strips of area w(x)Δxw(x)\Delta xw(x)Δx at depth s(x)s(x)s(x).
  • Moments and Centers of Mass
    • Balance Points: Finding the point where a distributed mass behaves as a concentrated point mass.

Visual Anchors

1. The "Slice, Sum, Integrate" Workflow

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2. Hydrostatic Force Setup on a Vertical Plate

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[!NOTE] In the diagram above, the total force is accumulated by treating each Δxstripashavingconstantdepths(x).Theforceononestripis≈ρg⋅s(x)⋅w(x)Δx\Delta x strip as having constant depth s(x). The force on one strip is \approx \rho g \cdot s(x) \cdot w(x) \Delta xΔxstripashavingconstantdepths(x).Theforceononestripis≈ρg⋅s(x)⋅w(x)Δx.

Definition-Example Pairs

  • Term: Linear Density Function
    • Definition: The mass per unit length of a 1D object, varying by position.
    • Real-World Example: A baseball bat is heavier near the barrel than at the handle; its linear density ρ(x)\rho(x)ρ(x) increases as xxx moves toward the barrel end.
  • Term: Pascal's Principle
    • Definition: The fluid pressure at a given depth is identical in all directions.
    • Real-World Example: When swimming deeply, the pressure on your eardrums is the same whether you face upward, downward, or sideways.
  • Term: Center of Mass
    • Definition: The mathematical point where the entire mass of a system can be balanced.
    • Real-World Example: Placing your finger exactly under the 50cm mark of a uniform meter stick so it rests perfectly horizontal without tipping.

Worked Examples

▶Example 1: Mass of a 1D Rod with Variable Density

Problem: A thin rod is oriented on the xxx-axis from x=0x = 0x=0 to x=4meters.Itslineardensityfunctionisgivenbyρ(x)=2x+1x = 4 meters. Its linear density function is given by \rho(x) = 2x + 1x=4meters.Itslineardensityfunctionisgivenbyρ(x)=2x+1 kg/m. Find the total mass of the rod.

Step-by-Step Solution:

  1. Identify the formula: The total mass is the integral of the density function over the length of the rod. m=∫abρ(x)dxm = \int_a^b \rho(x) dxm=∫ab​ρ(x)dx
  2. Substitute the given values: m=∫04(2x+1)dxm = \int_0^4 (2x + 1) dxm=∫04​(2x+1)dx
  3. Find the antiderivative: ∫(2x+1)dx=x2+x\int (2x + 1) dx = x^2 + x∫(2x+1)dx=x2+x
  4. Evaluate the definite integral: m=[x2+x]04m = [x^2 + x]_0^4m=[x2+x]04​ m=(42+4)−(02+0)=16+4=20m = (4^2 + 4) - (0^2 + 0) = 16 + 4 = 20m=(42+4)−(02+0)=16+4=20
  5. Final Answer: The total mass of the rod is 20 kg.
▶Example 2: Setting up a Hydrostatic Force Integral

Problem: A rectangular dam is 10 meters wide and extends 5 meters vertically downward into the water. Set up the integral to find the hydrostatic force against the dam, assuming the top of the dam is at the water's surface (x=0).Lettheweightdensityofwaterbeρgx=0). Let the weight density of water be \rho gx=0).Lettheweightdensityofwaterbeρg.

Step-by-Step Solution:

  1. Define the coordinate system: Let the x−axispointdownward,withx=0x-axis point downward, with x = 0x−axispointdownward,withx=0 at the surface.
  2. Determine depth s(x)s(x)s(x): Because x=0isthesurface,depthissimplys(x)=xx=0 is the surface, depth is simply s(x) = xx=0isthesurface,depthissimplys(x)=x.
  3. **Determine width w(x):∗∗Thedamisrectangular,sowidthisconstantatw(x)=10w(x):** The dam is rectangular, so width is constant at w(x) = 10w(x):∗∗Thedamisrectangular,sowidthisconstantatw(x)=10.
  4. Identify integration bounds: The dam goes from the surface to 5 meters deep, so x∈[0,5]x \in [0, 5]x∈[0,5].
  5. Construct the integral: F=∫abρg⋅s(x)⋅w(x)dxF = \int_a^b \rho g \cdot s(x) \cdot w(x) dxF=∫ab​ρg⋅s(x)⋅w(x)dx F=∫05ρg⋅(x)⋅(10)dxF = \int_0^5 \rho g \cdot (x) \cdot (10) dxF=∫05​ρg⋅(x)⋅(10)dx F=10ρg∫05xdxF = 10\rho g \int_0^5 x dxF=10ρg∫05​xdx

Checkpoint Questions

  1. If a rod has a constant density of 5 kg/m and a length of 3 m, why do we not need integration to find its mass?
  2. According to Pascal's Principle, if a vertical plate is rotated horizontally at the exact same depth, how does the pressure on it change?
  3. When calculating the hydrostatic force on a vertical submerged plate, what does the expression w(x)Δxw(x)\Delta xw(x)Δx physically represent in the Riemann sum?
  4. Why is the center of mass colloquially referred to as the "balancing point" of a system?
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