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Integrate a density profile before evaluating pressure and buoyancy

Hydrostatic force balance on a thin fluid layer gives dp/dz = ρg when depth z increases downward. The familiar pressure increment ρgh assumes constant density. Here the stipulated stable profile is ρ(z) = 1000(1 + βz) kg/m³ with nonnegative β.

Learning outcome: Integrate pressure, compute opposite-face forces and independently integrate the displaced-fluid weight. Compare the uniform limit.

Starting recap

Derive equations with their initial or boundary conditions, integrate distributed profiles, and check model domains independently of numerical outputs.

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For β = 0, buoyancy becomes 50 N at every supported depth. For β = 0.2 and top depth 0, average density is 1050 kg/m³ and buoyancy is 52.5 N.

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Predict

At β = 0.2 and slab top depth 1 m, buoyancy is:

Your prediction and reason

Experiment

Integrate pressure, compute opposite-face forces and independently integrate the displaced-fluid weight. Compare the uniform limit.

Integrate depth-dependent density before using buoyancy

Worked teaching model · synthetic values · Stable stipulated liquid density rho(z)=1000(1+beta z) kg/m³, depth z positive downward,g=10. A fully submerged vertical slab has area.01 m²,height.5 m; top/bottom pressure forces set buoyancy. Atmospheric surface pressure is the gauge zero. No compressibility claim or physical immersion.

Density depth coefficient: 0.2 1/m Top depth of submerged slab: 1 m

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Integrated gauge-pressure profile

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: integral rho(z)g dz (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Integrated gauge-pressure profile

integral rho(z)g dz

depth below surface (m)	gauge pressure (Pa)
0	0
0.03125	313.4766
0.0625	628.9063
0.09375	946.2891
0.125	1265.625
0.15625	1586.914
0.1875	1910.156
0.21875	2235.352
0.25	2562.5
0.28125	2891.602
0.3125	3222.656
0.34375	3555.664
0.375	3890.625
0.40625	4227.539
0.4375	4566.406
0.46875	4907.227
0.5	5250
0.53125	5594.727
0.5625	5941.406
0.59375	6290.039
0.625	6640.625
0.65625	6993.164
0.6875	7347.656
0.71875	7704.102
0.75	8062.5
0.78125	8422.852
0.8125	8785.156
0.84375	9149.414
0.875	9515.625
0.90625	9883.789
0.9375	10253.91
0.96875	10625.98
1	11000
1.03125	11375.98
1.0625	11753.91
1.09375	12133.79
1.125	12515.63
1.15625	12899.41
1.1875	13285.16
1.21875	13672.85
1.25	14062.5
1.28125	14454.1
1.3125	14847.66
1.34375	15243.16
1.375	15640.63
1.40625	16040.04
1.4375	16441.41
1.46875	16844.73
1.5	17250
1.53125	17657.23
1.5625	18066.41
1.59375	18477.54
1.625	18890.63
1.65625	19305.66
1.6875	19722.66
1.71875	20141.6
1.75	20562.5
1.78125	20985.35
1.8125	21410.16
1.84375	21836.91
1.875	22265.63
1.90625	22696.29
1.9375	23128.91
1.96875	23563.48
2	24000
2.03125	24438.48
2.0625	24878.91
2.09375	25321.29
2.125	25765.63
2.15625	26211.91
2.1875	26660.16
2.21875	27110.35
2.25	27562.5
2.28125	28016.6
2.3125	28472.66
2.34375	28930.66
2.375	29390.63
2.40625	29852.54
2.4375	30316.41
2.46875	30782.23
2.5	31250

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Gauge pressure at slab top	11000	Pa
Gauge pressure at slab bottom	17250	Pa
Integrated upward buoyancy	62.5	N
Top-density times g times volume estimate	60	N
Displaced-fluid average density	1250	kg/m³

Buoyancy equals displaced-fluid weight even when density varies, but one endpoint density is not its volume average.

With downward depth z, dp/dz=rho(z)g. Integrating gives p_gauge=rho0 g(z+beta z²/2). For the vertical slab, net pressure force is area[p(bottom)−p(top)], equal to integral rho g dV. The linear-density volume average is at middepth. Side pressure forces cancel horizontally in this ideal symmetric geometry.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Buoyancy equals displaced-fluid weight even when density varies, but one endpoint density is not its volume average.

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Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Integrate a density profile before evaluating pressure and buoyancy

Integrating from the surface gives gauge pressure 10000(z + βz²/2) Pa for g = 10. Atmospheric pressure is the common reference and cancels from opposite-side net forces; it does not vanish as an absolute pressure. The density profile is supplied independently rather than inferred from a claimed universal liquid equation of state.

A vertical submerged slab has horizontal area 0.01 m² and height 0.5 m. Its upward pressure force is area times the bottom-minus-top pressure. At β = 0.2 m⁻¹ and top depth 1 m, top gauge pressure is 11000 Pa and bottom pressure is 17250 Pa, giving buoyancy 62.5 N.

The displaced-fluid average density is 1250 kg/m³, its midpoint density for this linear profile. Multiplying that by displaced volume 0.005 m³ and g gives the same force. Using the top density alone gives 60 N and misses the deeper fluid’s greater weight. Archimedes’ displaced-weight statement still holds; the averaging must match the actual volume.

Changed-case check

Predict before reading the comparison. For β = 0, buoyancy becomes 50 N at every supported depth. For β = 0.2 and top depth 0, average density is 1050 kg/m³ and buoyancy is 52.5 N.

Retained investigation

Integrate pressure, compute opposite-face forces and independently integrate the displaced-fluid weight. Compare the uniform limit. Why does a common atmospheric reference cancel from buoyancy but not from an absolute pressure report?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All prose, functions and evidence packets are original. Bounded mathematical models do not establish measured apparatus, supervised practical skill, independent learner mastery or institutional credit. No physical loading, liquids, heat or sound exposure is requested.

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Why does a common atmospheric reference cancel from buoyancy but not from an absolute pressure report?

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