Build a physical integral from local force

Try

A spring requires greater pulling force as it stretches. Multiplying the final force by the entire displacement would treat that maximum force as if it acted throughout. Predict why integrating the varying force gives a smaller work value for a linear spring starting at zero extension.

Predict

Pressure in a stationary liquid increases with depth. Predict whether a tall vertical plate can always be analyzed by multiplying its top-edge pressure by its entire area. Then consider a plate whose width changes with depth: both pressure and strip area may vary.

Experiment

Keep the local quantity and the accumulated quantity distinct.

Application	Local contribution	Units
One-dimensional work	F(x) dx	newton-metres, or joules
Hydrostatic force	γh·w(h) dh	newtons
Pumping work	weight of slice × lift distance	joules

For a spring F(x)=20x newtons, work from zero to 1/2 metre is ∫20x dx=5/2 joules. The final force is 10 newtons; multiplying it by 1/2 metre would give 5 joules, twice the correct value. Draw the triangular area under the force graph to verify the result.

Understand

For motion along one coordinate, signed work is ∫F(x)dx, with F the force component in the direction of increasing x. If force opposes the directed displacement, its contribution is negative. For F(x)=3−3x from 0 to 2, positive and negative triangular contributions cancel to zero net work. Integrating |F| gives 3 joules of summed magnitudes, a different quantity. More general motion requires force dotted with displacement, developed in vector calculus.

Hydrostatic gauge pressure at depth h in a uniform stationary liquid is γh, where γ is weight density in force per volume. If mass density ρ is supplied instead, γ=ρg with an appropriate gravitational acceleration. Confusing those densities produces a missing or duplicated factor of g. Uniform atmospheric pressure cancels only when the problem concerns the net gauge contribution under the stated exposure assumptions.

A horizontal strip of a vertical plate has area w(h)dh, so force is ∫γh w(h)dh. For width 2 metres over depths 1 to 3 and γ=10 newtons per cubic metre, force is 20∫₁³h dh=80 newtons. These deliberately simple values are synthetic teaching assumptions, not calibration for a particular real fluid. For a triangular plate with w(h)=h from depth zero to 2, the force becomes γ∫₀²h²dh=8γ/3 in the consistent units.

For pumping, choose a vertical coordinate y and derive both slice volume and lift distance. A full cylindrical tank with base area 3 square metres and height 4 metres, weight density 10 newtons per cubic metre, has slice weight 30dy. Pumping just to the top requires distance 4−y, giving work 30∫₀⁴(4−y)dy=240 joules. Pump inefficiency, flow losses and extra outlet height would change the model and must be added explicitly if required.

These integrals require a physical model before calculus begins. A correct primitive cannot fix a mistaken strip width, a distance measured from the wrong reference, or incompatible units. Mark the origin, depth direction and outlet level on a sketch so the formula represents the intended situation.

Apply

Write the local force or slice weight first, then multiply by the relevant displacement or strip area. In written tasks, state the uniform-density and idealized-loss assumptions. Distinguish signed net work from magnitude accumulation and distinguish pressure from total force.

Continue

This completes a series-and-applications unit, not an independent guarantee of BC examination readiness or a full semester credential. Further differential equations, cumulative unfamiliar assessments and source-by-source objective review remain necessary parts of the broader program.