Lesson 19: Applications: Work, Fluid Pressure, and Centers of Mass

Integrals compute more than areas. This lesson covers three physical applications where the "slice, approximate, integrate" pattern wears different clothes: work (force varying along a path), fluid pressure (force from a liquid against a vertical surface), and centres of mass (balance points of regions). Each has a trap that catches the unprepared, and each trap is a modelling question, not a calculus one: what exactly is a single slice, and what does the integral accumulate?

Work: when force varies

Work is force × distance only for constant force. For a varying force F(x) along a line, slice the path: the slice at x of width dx needs work ≈ F(x) dx, and W = ∫ F(x) dx. Worked Example 1 (Hooke's law). A spring exerts F = kx. If 20 N stretches it 0.05 m, then k = 20/0.05 = 400 N/m. The work to stretch it from natural length to 0.15 m: W = ∫ from 0 to 0.15 of 400x dx = 200 x^2 evaluated = 200(0.0225) = 4.5 J. Note the force at full stretch is 60 N, but the work is not 60 × 0.15 = 9 J — the force grew from 0, averaging 30 N, and 30 × 0.15 = 4.5 ✓. The integral is the honest average.

Pumping problems: every slice travels a different distance

Worked Example 2. A cylindrical tank (radius 2 m, height 3 m) is full of water (weight density ρg = 9800 N/m³). Find the work to pump all the water out of the top. Slice horizontally at height y (measured from the bottom), thickness dy: the slice has volume π(2^2) dy = 4π dy, weight 9800·4π dy newtons, and must be lifted (3 − y) metres:

W = ∫ from 0 to 3 of 9800·4π·(3 − y) dy = 39200π [ 3y − y^2/2 ] from 0 to 3
  = 39200π (9 − 4.5) = 176400π ≈ 5.54 × 10^5 J

The (3 − y) factor is the entire problem: top slices barely travel, bottom slices travel nearly 3 m. Set the coordinate system first, express lift distance in it, and the rest is arithmetic. (Had the tank been only half full, the limits — not the integrand's shape — would change: slice y from 0 to 1.5.)

Fluid force: pressure grows with depth

Pressure at depth h is ρgh (gauge), and force on a surface is pressure × area — but pressure varies down a vertical plate, so integrate strip by strip. Worked Example 3. A dam face is 20 m wide with water 5 m deep against it. A horizontal strip at depth h, thickness dh, has area 20 dh and pressure 9800h:

F = ∫ from 0 to 5 of 9800·20·h dh = 196000 · (25/2) = 2.45 × 10^6 N

Using the surface pressure (0) or the bottom pressure (49000 Pa × 100 m² = 4.9 × 10^6) would both be wrong; the integral's answer equals the pressure at the centroid depth (2.5 m) times the area — a general theorem (F = ρg·(depth of centroid)·A) that slicing proves and then lets you use.

Centres of mass: balance as an average position

For a region under y = f(x) on [a, b] (uniform density), the centroid is the area-weighted average position: x̄ = (1/A) ∫ x f(x) dx and ȳ = (1/A)·(1/2) ∫ f(x)^2 dx — the ȳ formula slices the region into vertical strips, each strip's own centre at height f(x)/2. Worked Example 4. Region under y = 4 − x^2 above the x-axis (so −2 ≤ x ≤ 2). Symmetry gives x̄ = 0 immediately — never compute what symmetry hands you. Area A = ∫ from −2 to 2 of (4 − x^2) dx = [4x − x^3/3] = (8 − 8/3) − (−8 + 8/3) = 32/3. Then ∫ f^2 dx = ∫ (16 − 8x^2 + x^4) dx from −2 to 2 = 512/15 (compute via [16x − 8x^3/3 + x^5/5]: 2(32 − 64/3 + 32/5) = 512/15), so ȳ = (1/(2A))·(512/15) = (512/15)/(64/3) = 8/5 = 1.6. The balance point (0, 1.6) sits below the peak (0, 4) — the region is bottom-heavy relative to its silhouette, as a balance point should be.

Common mistakes

• Measuring pumping lift from the wrong reference. The lift distance is from the slice's position to the exit point, (3 − y) here. Using y (distance from the bottom) reverses the effort profile and the answer. Draw the tank, mark the exit, write the distance in the drawing's coordinates.
• Using mass density where weight density belongs. With ρ = 1000 kg/m³, the weight of a slice is ρ·g·volume = 9800 × volume newtons. Dropping g understates every work/force answer by a factor of 9.8. Units audit catches this: joules are newton-metres, so a force must appear in newtons.
• Constant-pressure shortcut on vertical plates. Force = pressure × area holds only for the centroid-depth pressure. Using surface or bottom pressure across the whole plate is the error Worked Example 3 exists to prevent.
• Computing x̄ when symmetry decides it. Symmetric region about x = 0 means x̄ = 0 by definition of balance. Examiners award the observation; grinders waste five minutes confirming it. (The reverse error — assuming symmetry that is not there — is equally real: check the limits and the function.)

Exam Experience

1. A spring requires a force of 30 N to hold it stretched 0.1 m beyond natural length. The work done stretching it from natural length to 0.2 m is

• (A) 3 J
• (B) 6 J
• (C) 12 J
• (D) 1.5 J

Answer: (B). k = 30/0.1 = 300 N/m; W = ∫ from 0 to 0.2 of 300x dx = 150(0.04) = 6 J. Choice (A) is 30 × 0.1-style reasoning applied to the wrong distance; constant-force thinking is the trap.

2. The x-coordinate of the centroid of the region under y = x^2 on [0, 1] (above the x-axis) is

• (A) 1/2
• (B) 2/3
• (C) 3/4
• (D) 3/5

Answer: (C). A = 1/3; ∫ from 0 to 1 of x·x^2 dx = 1/4; x̄ = (1/4)/(1/3) = 3/4. The centroid leans right because the region's area concentrates at larger x — intuition and arithmetic agree.

3. The gauge pressure at depth 4 m in water (weight density 9800 N/m³) is

• (A) 9800 Pa
• (B) 19600 Pa
• (C) 39200 Pa
• (D) 49000 Pa

Answer: (C). P = ρgh = 9800 × 4 = 39200 Pa. (D) is the pressure at 5 m — depth bookkeeping is the tested skill.

Free-response question. A cylindrical tank has radius 3 m and height 4 m and stands full of water (weight density 9800 N/m³). (a) Set up the integral for the work required to pump all the water to the top edge of the tank, defining your coordinate. (b) Evaluate the work. (c) The tank is instead only half full (depth 2 m). Without recomputing from scratch, explain how the set-up changes and give the new work value.

Scoring guideline.

• (a) 1 point. With y measured from the bottom, a slice at height y has volume 9π dy, weight 9800·9π dy = 88200π dy, lift (4 − y): W = ∫ from 0 to 4 of 88200π (4 − y) dy. Any consistent coordinate with the correct lift expression earns the point.
• (b) 1 point. 88200π [4y − y^2/2] from 0 to 4 = 88200π (16 − 8) = 705600π ≈ 2.22 × 10^6 J.
• (c) 2 points. 1 point for the explanation: only the limits change — slices run y = 0 to 2 (water occupies the bottom 2 m), same integrand. 1 point for the value: 88200π [4y − y^2/2] from 0 to 2 = 88200π (8 − 2) = 529200π ≈ 1.66 × 10^6 J — note it is more than half the full-tank work, because the water that remains is the deep water with the longest lifts. That observation, or the correct value with working, completes the point.

Going deeper: a semicircular centroid, a torus, and a rope

The centroid machinery of Worked Example 4 scales to curved boundaries. For the semicircular region under y = sqrt(1 − x^2) on [−1, 1]: symmetry gives x̄ = 0; the area is A = π/2; ∫ f(x)^2 dx = ∫ from −1 to 1 of (1 − x^2) dx = 4/3; hence ȳ = (1/(2A))·(4/3) = 4/(3π) ≈ 0.424 — one of the classical constants of the subject, obtained with no new ideas, just the strip formula applied carefully. Pappus's centroid theorem then converts centroids into volumes for free: rotating a plane region about an external axis, the volume equals (area) × (distance travelled by the centroid). A disc of radius 1 centred at (2, 0), rotated about the y-axis, makes a torus: V = π · (2π·2) = 4π^2 ≈ 39.5. Lesson 10's slicing proves this in general; here it pays off a result that would otherwise need washers in two pieces. Finally, the rope problem, pumping's closest relative: a uniform rope 20 m long hangs from a roof, weighing 9.8 N per metre. Lifting it to the roof: when y metres have been raised, 20 − y metres still hang, weighing 9.8(20 − y) N, and raising the next dy metres costs that weight × dy:

W = ∫ from 0 to 20 of 9.8(20 − y) dy = 9.8 · 200 = 1960 J

The integrand decreases as the job proceeds — the load literally lightens — and recognising *which* quantity shrinks (rope length, not height) is the modelling step. Work problems are won or lost there, never in the integration.