Area, volume, averages and physical accumulation

Decide what one small piece contributes

A flow of 4 litres per minute lasting half a minute adds 2 litres. A slice with cross-sectional area 4 square metres and thickness half a metre has volume 2 cubic metres. Both calculations multiply a quantity per unit input by a small input width, but their meanings and units differ.

For a changing rate or cross-section, divide the input interval into short pieces. Use a representative value on each piece, multiply by its width, and add. A definite integral is the limit of these sums when the largest width tends to zero. First identify the piece you need; the integral sign records the accumulation after the model is chosen.

This lesson uses ordinary single-variable integrals on finite intervals. Continuous functions are sufficient for the calculations below; a bounded function with finitely many jump discontinuities can also be integrated piece by piece. An infinite interval or an unbounded integrand requires an improper-limit and convergence check, rather than an automatic endpoint substitution.

Useful preparation is functions and their domains, signed fractions, basic area and volume, and interpreting an antiderivative. The preparation links recall signed finite accumulation and the endpoint rule. On a first reading, follow the signed-area, cone, average and tank examples. Return to the shell comparison, weighted centre, pumping model and curve-length extension after those examples.

Only the stated example and finite interval are shown. The exact adjacent values provide a text equivalent; the calculations, rather than pixel areas, establish the integrals.

Cancellation answers a different question from area

Take f(x)=x−1 from x=0 to x=3, with both plotted coordinates in the same length unit. Between 0 and 1, the graph lies below the horizontal axis and forms a triangle of area 1/2. Between 1 and 3 it lies above the axis and forms a triangle of area 2. The forward signed integral is −1/2+2=3/2 square units; the geometric area is 1/2+2=5/2 square units.

For continuous f on a finite interval [a,b] with a<b, geometric area between its graph and the horizontal axis is ∫ₐᵇ|f(x)|dx. The signed integral is ∫ₐᵇf(x)dx. In general, the absolute value of the final signed answer is not the area: |3/2| is still 3/2, not 5/2. Split at every relevant sign change before adding geometric pieces.

If F′=f throughout an interval containing [a,b] and f is continuous there, the fundamental theorem gives ∫ₐᵇf(x)dx=F(b)−F(a). Here F=x²/2−x, so F(3)−F(0)=3/2. This endpoint subtraction computes the signed integral; it does not remove cancellation. Reversing endpoints negates the integral. A geometric region has a nonnegative area regardless of which direction you describe it.

Before using the table, predict whether a final absolute value can recover both geometric pieces.

Solid blue line: y=x−1. Hatched orange: the part below zero; blue shading: the part above zero. The dashed line marks the sign change at x=1.

A strip reaches from its lower boundary to its upper boundary

For continuous functions f and g on [a,b], the vertical strip between their graphs has height |f(x)−g(x)|. Its geometric area is therefore ∫ₐᵇ|f−g|dx. If f≥g throughout the interval, the absolute value becomes f−g. “Upper minus lower” concerns vertical position; both graphs can be below zero.

Compare y=x and y=x² from 0 to 2. Their intersections solve x=x², giving x=0 and x=1. On (0,1), x is above x²; on (1,2), x² is above x. Integrating x−x² gives 1/6 on [0,1], whereas integrating x²−x gives 5/6 on [1,2]. Add them to obtain area 1 square unit. Keeping x−x² over the whole interval instead gives the signed difference −2/3.

A sketch helps find the intended boundaries, but intersections and ordering need algebra or another all-input argument. Check interval endpoints as well as intersections. An equality without an order change need not require a new formula. With many crossings, the absolute-value integral is still the statement of the area; our finite splitting method applies when the relevant changes can be partitioned into finitely many pieces.

Horizontal strips can be easier. If the region at each height y runs continuously from x=L(y) to x=R(y), with R≥L and continuous boundaries on [c,d], then area is ∫꜀ᵈ[R(y)−L(y)]dy. Use the actual horizontal extent. A strip that meets two separated pieces requires their two widths, rather than filling the gap.

Predict which curve is higher at x=1/2 and at x=3/2 before choosing a subtraction order.

Solid blue: y=x. Dashed orange: y=x². The vertical dotted line marks the interior intersection. Shading follows the actual region between them.

A slice contributes area times thickness

Let A(x) be the actual area of the complete cross-section of a solid in the plane perpendicular to its x-direction. The solid extends from a to b, with a<b. When A is a nonnegative continuous function, the volumes of thin slices converge to V=∫ₐᵇA(x)dx. Each cross-section counts all occupied pieces of that plane once. Multiplying an outline width by thickness would give area, not volume.

Consider a right circular cone of height 4 metres and base radius 3 metres, with x measured from its tip along its axis. Similar triangles give radius r(x)=3x/4 for 0≤x≤4, when x and r are numerical lengths in metres. Its perpendicular cross-section is a disk, so A(x)=9πx²/16 square metres. Thus V=(9π/16)[x³/3]₀⁴=12π cubic metres.

The same construction works for noncircular slices. If a solid has square cross-sections with side s(x), use A=s²; for triangular sections use the area appropriate to their actual base and height. A drawing of one outside boundary alone does not establish the cross-section's shape. The spatial geometry supplies A before calculus adds the slices.

Solid boundary: upper generator r=3x/4. Dashed boundary: its reflected lower edge in this axial section. The marked slice is at x=2 m; its disk radius is 3/2 m. This axial drawing is not a perspective view of a disk.

A washer subtracts two disk areas

Rotate the region between y=x² and y=x on 0≤x≤1 around the horizontal axis y=0. At a fixed x, the vertical segment runs from x² to x. Its rotation is a washer: an outer disk of radius R=x with a concentric inner disk of radius r=x² removed. The cross-sectional area is π(R²−r²)=π(x²−x⁴), not π(R−r)².

The volume is π∫₀¹(x²−x⁴)dx=π(1/3−1/5)=2π/15 cubic units. At x=1/2, R=1/2 and r=1/4, so the washer area is 3π/16 square units. Subtracting radii first would instead give π/16 and describe a different disk.

In general, use V=π∫ₐᵇ[R(x)²−r(x)²]dx when the perpendicular cross-sections really are washers and 0≤r≤R are continuous. A disk is the case r=0. Radii are nonnegative distances from the stated rotation axis. About y=k, a boundary's distance is |y−k|; its algebraic height is not automatically its radius.

If a vertical region crosses the axis, its rotation can fill a disk: the segments on opposite sides overlap after rotation. Do not subtract a hole that the region itself fills, or add the two sides as separate volumes. Determine the actual cross-section and split the interval wherever its description changes.

Solid blue circle: outer radius R. Dashed orange circle: inner radius r. Hollow centre: the removed inner disk. Both radial segments start at the same rotation-axis point.

A different slicing direction can describe the same solid

Rotate the same region, y=x² to y=x for 0≤x≤1, about the vertical axis x=0 instead. A thin vertical strip at x sweeps out a cylindrical shell. Its radius is x, its height is x−x² and its radial thickness is Δx. The leading small-volume contribution is circumference times height times thickness: 2πx(x−x²)Δx.

Hence V=2π∫₀¹(x²−x³)dx=2π(1/3−1/4)=π/6 cubic units. The shell approximation becomes exact only in the limit; a shell of finite thickness has an additional thickness-squared contribution. Here the region is wholly on one side of the axis, so distinct radial strips do not describe the same shell twice.

Check this result with horizontal washers. At 0≤y≤1, the region runs from x=y to x=√y. Rotation about x=0 gives outer radius √y and inner radius y. Thus V=π∫₀¹(y−y²)dy=π(1/2−1/3)=π/6. Both integrals measure the same solid. The earlier value 2π/15 used another axis and therefore another solid.

For a continuous nonnegative height h(x) on a one-sided radial interval 0≤a<b about x=0, shells give V=2π∫ₐᵇx h(x)dx. A shifted or opposite-side region needs the corresponding positive radial distance and matching limits. If a region occupies both sides of an axis, equal-radius strips can overlap in the swept solid; naïvely summing both shell volumes double counts the overlap. Choose a nonoverlapping description or use actual cross-sections.

Solid blue: y=x. Dashed orange: y=x². Thick green vertical segment: one shell's height. Dotted pink horizontal segment: one washer's radial interval. The vertical axis is the rotation axis; the segments are at different representative levels.

An average preserves the same total over the same interval

For an integrable function f on [a,b] with a<b, its uniform average is f̄=(1/(b−a))∫ₐᵇf(x)dx. A constant height f̄ across the interval has the same signed integral as f. This mean weights equal input lengths equally. It is not generally the mean of the two endpoint values or of a few irregularly spaced samples.

For f(x)=x² on [0,3], the integral is [x³/3]₀³=9 and the interval width is 3, so f̄=3. The endpoint mean is (0+9)/2=9/2, which is different. The horizontal line at height 3 gives a rectangle of area 9, matching the original area. Signed means are also valid: positive and negative function values can cancel.

The mean has the units of f because the integration unit is divided out. An average velocity over a time interval is displacement divided by elapsed time. Average speed is distance divided by elapsed time. If f is continuous, its minimum and maximum bound the average, and the intermediate value theorem supplies some c∈[a,b] with f(c)=f̄. For x² here c=√3. Without continuity, a mean need not be an attained value: a function taking only 0 and 1 on two half-intervals has mean 1/2.

A zero-length interval gives division by zero in this definition. For piecewise constant values, multiply each value by the length for which it applies before dividing by the total length. A rate of 2 for one minute and 4 for two minutes has time average (2·1+4·2)/3=10/3, not 3.

Solid blue: f=x². Dashed orange: uniform mean 3. The marked intersection is at x=√3. Each total is 9 on the same interval.

Name what receives equal weight

A weight w(x)≥0 can express how much mass, volume or other importance belongs to each small input interval. If w and fw are integrable and W=∫ₐᵇw(x)dx>0, the weighted mean is (∫ₐᵇf(x)w(x)dx)/W. The denominator is total weight, not automatically b−a. Nonnegative weights keep the result between bounds of f; arbitrary signed weights do not have that averaging interpretation.

On [0,1], the uniform mean of f(x)=x is 1/2. With w(x)=2x, total weight is ∫₀¹2x dx=1 and weighted mean is ∫₀¹2x²dx=2/3. Larger inputs receive more weight. The same values and interval can therefore have different means when the question changes what is weighted equally.

For a thin rod, linear mass density λ is mass per unit length. A short piece has mass approximately λ(x)Δx, so total mass is M=∫λ(x)dx with λ≥0. Take a 2-metre rod with density λ=(1 kg/m)+(1 kg/m²)x. When x is the numerical position in metres, the numerical density in kg/m is 1+x. Its mass is ∫₀²(1+x)dx=4 kg.

Its centre of mass along the rod is x̄=(∫xλ(x)dx)/M when M>0. The numerator, called the first moment, has units kg·m: ∫₀²x(1+x)dx=14/3 kg·m. Dividing by 4 kg gives 7/6 m, to the right of the midpoint because the density increases. This is an ideal thin-rod model. For a three-dimensional material with density varying across a slice, its full slice mass, rather than one centre-point density, must be used.

Solid blue: linear density 1+x in numerical metre coordinates. Dotted grey: equal half-metre divisions. The triangle marks the geometric midpoint; the filled circle marks the density-weighted centre of mass.

A rate determines change; the initial amount determines the total

If a quantity Q has continuous net rate Q′(t)=r(t), then Q(t)=Q(t₀)+∫ₜ₀ᵗr(s)ds. This is the fundamental theorem with an initial value. The symbol s is a temporary integration variable; the upper endpoint t is the time at which you want the total. Rate multiplied by time has the units of Q.

An ideal tank initially contains 5 litres. During the next 3 minutes, inflow is 4 L/min and outflow is (2 L/min²)t. Net rate is r(t)=4−2t in numerical L/min when t is in minutes. Therefore Q(t)=5+4t−t² litres. At 3 minutes it contains 8 litres, a net increase of 3 litres. Inflow alone would add 12 litres, but 9 litres have left.

The net rate is positive before t=2, zero at t=2 and negative after it. The amount consequently reaches its maximum 9 litres at 2 minutes. A zero rate here means the amount momentarily stops increasing; it does not mean the tank is empty. The model requires capacity at least 9 litres, no other gains or losses, and the stated rates remaining valid. On this interval the amount never becomes negative. If a tank empties or overflows, continue only with an updated physical model.

For motion on a line, signed velocity v is the rate of position change. With v(t)=2−t m/s for 0≤t≤3 seconds, displacement is ∫₀³(2−t)dt=3/2 m. Speed is |v|, so distance is ∫₀²(2−t)dt+∫₂³(t−2)dt=5/2 m. Average velocity is 1/2 m/s; average speed is 5/6 m/s. Reversing direction changes the sign of velocity, while travelled distance keeps increasing.

Upper solid blue: net rate 4−2t. Lower dashed orange: amount 5+4t−t². Dotted vertical guides mark the same t=2 min. Each panel has its own labelled vertical scale.

Work adds force along a displacement, not force alone

A force is a push or pull, measured in newtons (N). Along a straight path, resolve the chosen force into its signed component F parallel to a coordinate x. Its work from a to b is W=∫ₐᵇF(x)dx, when that force component is a continuous function of position along this traversal. A constant force gives F(b−a). Work has units N·m, called joules (J); integrating force over time would instead have units N·s and describe impulse.

A positive work contribution means the force component and displacement point in the same direction; a negative contribution means they oppose. If position reverses, split the traversal or use ∫F(x(t))x′(t)dt with the actual time path. A force depending on history, time or speed cannot automatically be replaced by a single position-only function. This is work by the specified force, not automatically all the energy used by a device.

In an ideal linear spring, the restoring force is −kx, where x measures extension from its natural length and k>0 is the stiffness in N/m. A balancing external force during slow stretching is +kx when inertia and losses are neglected. With k=200 N/m, stretching from 0 to 0.1 m requires integrating 200x from x=0 to x=0.1. The endpoint difference of 100x² is 100·(0.1)²−100·0²=1 J. The spring's own restoring force does −1 J over that stretch.

The final external force is 20 N, but it was smaller earlier. Using 20 N times the whole 0.1 m gives 2 J and overestimates the work. Here the mean force is 10 N because the force grows linearly, so mean force times displacement gives 1 J. The linear spring relation is a model over its valid extension range, not a rule for every elastic material.

Solid blue: F=200x. Blue triangle: external work 1 J. Dashed orange rectangle: final force times full extension, 2 J. The spring restoring force has the opposite sign.

Different layers travel different distances

Consider a vertical cylindrical tank of constant cross-sectional area 2 m², containing water from height y=0 to y=3 m. Pump it to an outlet at height 4 m. Take constant mass density ρ=1000 kg/m³ and constant gravitational acceleration g=9.81 m/s². A layer of thickness Δy near height y has mass approximately ρ·2Δy and weight approximately ρg·2Δy newtons.

That layer must rise 4−y metres. Its ideal lifting work is therefore approximately ρg·2(4−y)Δy. Integrating all layers gives W=19620∫₀³(4−y)dy=19620[4y−y²/2]₀³=147150 J. The integration variable names the layer's original height, not the height of a single droplet throughout its later trip.

As a separate check, the water has mass 6000 kg and initial centre height 1.5 m. Every element finishes at height 4 m, so the centre-height rise is 2.5 m. The potential-energy increase 6000·9.81·2.5 is also 147150 J. This ideal calculation ignores friction, pump inefficiency and changes of kinetic energy; actual supplied energy needs those additional contributions. For a changing tank shape, replace 2 by its actual cross-sectional area A(y).

Blue shading: original water. Thick orange line: a representative layer. Green dashed segment: that layer's lifting distance. Tank width is schematic; its actual cross-sectional area is 2 m².

Optional: a sloping piece has length as well as horizontal width

For a graph y=f(x) with continuous derivative on [a,b] and a<b, a short chord has length √((Δx)²+(Δy)²). Factoring out positive Δx and taking the limit gives curve length L=∫ₐᵇ√(1+f′(x)²)dx. The mean value theorem relates each chord slope to an intermediate derivative; continuity makes the resulting Riemann sums converge. Both coordinates must be expressed in compatible length units. This is the length of the graph traced once as x increases.

The cone's generating line is y=3x/4 on [0,4], in metres. Its derivative is 3/4, so √(1+f′²)=5/4 and L=(5/4)·4=5 m. The same value follows from the 3–4–5 right triangle. Integrating y itself would measure area under the line, not its length.

Rotating a nonnegative continuously differentiable graph y=f(x) around y=0 gives lateral surface area S=∫ₐᵇ2πf(x)√(1+f′(x)²)dx: local circumference times sloping length. This formula covers the swept side surface, not automatically end caps. For the cone it gives 2π∫₀⁴(3x/4)(5/4)dx=15π m². The base adds a separate 9π m² if a closed cone's total surface is requested.

The surface-area formula describes a graph revolved once about its axis. More general parametrisations can retrace a surface and need multiplicity checks; corners can be handled in separate smooth pieces when the relevant lengths are finite. The later study of curves, surfaces and multiple integrals extends these ideas. No elementary primitive is guaranteed merely because a geometric formula has been set up correctly.

Solid blue: generating line of length 5 m. Dashed orange: axial length 4 m. Dotted green: base radius 3 m. Lateral area is 15π m²; the additional base area is 9π m².

Check the quantity before checking the antiderivative

Name the quantity you want, the input variable and its interval. Describe one representative strip, slice, shell, time step or moving layer. Multiply its factors with units: width times height, cross-section times thickness, rate times time, density times length, or force times displacement. Then decide which contributions are signed and which must remain nonnegative.

Check boundary ordering, axis distances, sign changes, initial amounts, positive averaging denominators and physical limits before integrating. Differentiate a proposed primitive and evaluate both endpoints. Compare units and bounds: a nonnegative volume cannot be negative, a nonnegative-weight mean cannot lie outside the values being averaged, and an accumulated amount must remain physically possible under its stated model.

Each graph retains its declared numerical scales; the cone's axial section and the 3–4–5 triangle use equal length scales in both coordinate directions. Exact tables and calculations establish the answers. Measured data need a justified interpolation or numerical integration method and an uncertainty estimate. Calculus can make a model's consequences exact without making uncertain measurements or idealised assumptions exact.

Worked examples

Try it yourself

1. For f=x−1 on [0,3], does |∫f| equal the geometric area?

   Show the answer and reasoning

   No: |∫f|=3/2, whereas ∫|f|=5/2.

   The negative triangle cancels part of the positive signed contribution before the final absolute value is taken. Geometric area adds both magnitudes.

2. On [0,2], the upper curve is y=−1 and the lower curve is y=−3. Find their enclosed strip area.

   Show the answer and reasoning

   4 square units.

   Upper minus lower is −1−(−3)=2. Integrating this positive height over width 2 gives 4; neither boundary has to be above zero.

3. A washer has outer radius 3 m and inner radius 1 m. Is its area π(3−1)²?

   Show the answer and reasoning

   No; its area is 8π m².

   Subtract disk areas: π·3²−π·1²=8π. Squaring the difference would describe a disk of radius 2 m.

4. A boundary at height y=5 m rotates about the line y=2 m. What is its radius?

   Show the answer and reasoning

   3 m.

   Radius is the distance |5−2|, not the absolute boundary height 5. The complete cross-section still determines whether that radius is outer or inner.

5. A solid extends from x=0 to x=2 m. Its perpendicular square cross-section has side x metres. Find its volume.

   Show the answer and reasoning

   8/3 m³.

   The area is x² m² in numerical coordinates. Integrate ∫₀²x²dx=8/3; integrating the side x would have the wrong units.

6. May you add shell volumes from both sides of an axis without checking overlap?

   Show the answer and reasoning

   No.

   Two strips at the same radius on opposite sides can sweep out overlapping shells. Count the actual swept region once or use nonoverlapping cross-sections.

7. A rate is 2 L/min for one minute and 4 L/min for two minutes. Find its time average.

   Show the answer and reasoning

   10/3 L/min.

   The total is 2·1+4·2=10 L over 3 min. An unweighted mean of the two displayed rates would wrongly give both durations equal importance.

8. Can the uniform mean formula be applied on [2,2]?

   Show the answer and reasoning

   No; its denominator is zero.

   The definite integral is zero, but division by the zero interval width is undefined. A limiting mean is a separate question with its own hypotheses.

9. For f=x on [0,1], find its mean with weight w=2x and compare it with the uniform mean.

   Show the answer and reasoning

   Weighted mean 2/3; uniform mean 1/2.

   The positive total weight is ∫₀¹2x dx=1, and the weighted numerator is ∫₀¹2x²dx=2/3. More weight lies toward larger x.

10. In the tank example, does r(2)=0 mean Q(2)=0?

    Show the answer and reasoning

    No; Q(2)=9 L.

    The rate describes change, whereas the quantity includes its initial amount and past accumulation. Here zero rate marks the transition from increasing to decreasing.

11. For v(t)=2−t m/s on [0,3] seconds, find displacement, distance and average speed.

    Show the answer and reasoning

    Displacement 3/2 m; distance 5/2 m; average speed 5/6 m/s.

    Velocity changes sign at 2 s. Add signed pieces for displacement and their positive magnitudes for distance, then divide distance by the 3 s elapsed time.

12. Why is 20 N times 0.1 m wrong for the slowly stretched spring with k=200 N/m?

    Show the answer and reasoning

    It treats the final force as constant; the correct external work is 1 J.

    The balancing force grows from 0 to 20 N. Its triangular force–extension area is half the 2 J endpoint rectangle.

13. Does integrating force in newtons over time in seconds give work in joules?

    Show the answer and reasoning

    No; it gives impulse in N·s.

    Work requires the force component along displacement, so its units are N·m. A time integral for work must also include velocity.

14. In the pumping example, does every layer travel 4 m?

    Show the answer and reasoning

    No; a layer starting at y rises 4−y metres.

    The bottom layer rises 4 m, the top layer 1 m. Using one 4 m distance for all layers would overestimate the ideal work.

15. The cone has radius 3 m, height 4 m and slant length 5 m. Find its lateral and closed total surface areas.

    Show the answer and reasoning

    Lateral area 15π m²; including its base gives 24π m².

    The surface-of-revolution integral gives only the side, π·3·5. The separate circular base has area 9π m².

16. Can a continuous finite-interval endpoint rule alone justify integrating a density 1/x from 0 to 1?

    Show the answer and reasoning

    No.

    This density is unbounded near 0 and undefined there. The improper limit ∫ε¹1/x dx=−ln ε diverges as ε tends to 0, so this model has infinite mass on that interval.

What to carry forward

• Choose the small physical or geometric contribution before choosing an antiderivative.
• Signed integrals allow cancellation; geometric area integrates a nonnegative width or height.
• Volume needs actual cross-sectional area or nonoverlapping radial shells, with radii measured from the chosen axis.
• A mean divides by the total input length or positive total weight appropriate to the question.
• An initial value plus accumulated net rate gives a total; absolute velocity gives distance.
• Density times length gives mass, and the force component along displacement gives signed work.
• Arc length, surface area, volume and work have different small contributions and different units.
• Correct calculus preserves the model's conditions; it does not establish an idealisation or remove measurement uncertainty.