Mass from a varying density, work as the integral of a varying force, Hooke's law and springs, the work of pumping a tank, and hydrostatic force on a vertical plate — all from the same representative-piece discipline.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Physical Applications
Objectives
Five outcomes. Each is an elementary product turned into an integral because one factor varies.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 594-609 — the section these objectives are drawn from
Warm-up
Physics gives mass as density times length, and work as force times distance. Both are products, and both assume the factors are constant.
Discussion prompt
A rod's density varies along its length. What replaces the product?
Hint: Take a piece short enough that the density barely changes.
Answer:
On a piece short enough, the density is effectively constant — so that piece's mass is the density there times its length. Adding the pieces is integrating.
\[ m = \int_{a}^{b}\rho(x)\,dx \]
That is the entire pattern of this section. Every formula is an elementary product, applied to a piece small enough for the varying factor to be constant, then integrated. What changes from problem to problem is which factor varies and how, not the method.
Concept
When a physical quantity is a product of factors and one of them varies with position, apply the product to a piece small enough for that factor to be constant, then integrate over the object's extent.
the representative piece — A piece of the object small enough that every varying quantity is effectively constant on it. Its contribution is an elementary product, and integrating adds the contributions.
\[ \text{quantity} = \int (\text{elementary product on a thin piece}) \]
This is the same discipline as Sections 6.1 through 6.4, with a physical quantity in place of a geometric one. Nothing about the calculus changes.
Figure (svg): When multiplication suffices, and when an integral is needed
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 690-700
Section
Section 1
Concept
A rod whose density varies has mass equal to the integral of that density along its length. A disk with radial density needs the ring's circumference as well.
linear density — Mass per unit length, which may vary with position. The mass of a short piece is the density there times its length, and integrating gives the total.
\[ m = \int_{a}^{b}\rho(x)\,dx \]
For a disk with density depending on the distance from the centre, the representative piece is a thin ring — and its area brings in a factor of the circumference, exactly as in Section 6.3's shells.
Figure (svg): Mass from a varying density: a thin piece has nearly constant density
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 690-698 — mass and density
Picture it
A rod whose density increases along it.
Figure (svg): Mass from a varying density: a thin piece has nearly constant density
The shaded area is a mass rather than an area, because the integrand's units are mass per length. What an integral means is decided entirely by its integrand's units.
Worked example
Example 6.25. The basic case.
\[ \text{A rod on } [0,4] \text{ has density } \rho(x)=1+0.8x \text{ kg/m. Find its mass.} \]
Identify the representative piece
Why: A short length of rod.
Write its mass
Why: Density there times its length.
\[ (1 + 0.8 x) \,dx \]
Set the limits
Why: The rod's extent.
\[ 0\text{ to } 4 \]
Integrate
Why: Term by term.
\[ 4 + 0.4(16) \]
State
Why: With units.
\[ 10.4 \text{kg} \]
Figure (svg): Mass from a varying density: a thin piece has nearly constant density
\[ m = \int_{0}^{4}(1+0.8x)\,dx = 10.4 \]
Verify: check against a constant-density estimate
Why: The density runs from 1 to 4.2 kg per metre, averaging 2.6 over a 4-metre rod — which gives 10.4 kg exactly, because the density is linear and a linear function's average is the average of its endpoint values. For a non-linear density the check would only be approximate, but it always bounds the answer between the minimum and maximum densities times the length.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 693-695
Fill the middle
A thin ring at radius r.
Fill in the blanks
dA = 2\pi\,r\,dr
Why: The ring's area is its circumference times its width, exactly as a shell's surface was in Section 6.3. Omitting the factor treats a two-dimensional object as a one-dimensional one.
Worked example
Checkpoint 6.25. The piece is a ring.
\[ \text{A disk of radius } 4 \text{ has density } \rho(r)=\sqrt{r} \text{ kg/m}^{2}. \text{ Find its mass.} \]
Identify the representative piece
Why: Density depends on the radius.
Write the ring's area
Why: Circumference times width.
\[ 2 \pi r \,dr \]
Write its mass
Why: Density times area.
\[ \sqrt{r} \times 2 \pi r \,dr \]
Simplify
Why: Combine the powers.
\[ 2 \pi r ^{\frac{3}{2}} \,dr \]
Integrate from 0 to 4
Why: The power rule.
\[ 2 \pi(\frac{2}{5}) (32) = 128 \pi / 5 \]
Figure (svg): The solution to Worked example a disk with radial density shown as a ladder of expressions, one row per legal move
\[ m = 2\pi\int_{0}^{4}r^{3/2}dr = \frac{128\pi}{5} \]
Verify: notice the shell factor reappearing
Why: The circumference factor is exactly Section 6.3's shell radius, appearing here because the object has circular symmetry and the density depends on the radius. As a check, the density averages somewhere between 0 and 2 kilograms per square metre over a disk of area 16 pi, about 50 square metres — so a mass of 80 kilograms is plausible, weighted toward the outside where both the density and the ring area are larger.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 695-697
Trap
\[ m = \int_{0}^{4}\sqrt{r}\,dr \]
Integrate the density against the radius alone
Why: The student treats the ring like a rod piece.
A ring's area is its circumference times its width, so the factor of twice pi r is missing and the answer is far too small.
\[ m = \int_{0}^{4}\sqrt{r}\cdot 2\pi r\,dr \]
Write the piece's area, then multiply by the density
Why: For a two-dimensional object the density is per unit area.
Units settle it: a density in kilograms per square metre must multiply an area, and integrating it against a length alone leaves kilograms per metre. Checking the units of the integrand against the differential catches this immediately.
Sorting
It depends on how the density varies.
Sort into buckets
Sort each object.
The rule is that the piece should be the set of points where the varying quantity is constant. For a radial density that set is a circle, which is why the ring's circumference appears.
Two truths and a lie
All three are about density.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. An integral represents an area only when its integrand is a height; when the integrand is a density it represents a mass, when it is a force a work, and so on. What the integral means comes entirely from the units of what is being integrated.
Prediction
Commit before reasoning.
Predict first
What makes a good representative piece?
Correct: One on which the varying quantities are constant.
\[ \text{piece} = \text{a level set of the varying quantity} \]
Why: The whole point of the piece is that the elementary product applies to it, and that requires the varying factor not to change across it. For a radial density that means a ring rather than a rectangle, because a ring is exactly the set of points at one radius. Choosing the piece to match the symmetry of the varying quantity is the single decision that makes these problems tractable.
Section
Section 2
Concept
When the force varies with position, work is the integral of force over distance. For a spring the force is proportional to the extension, so the work is quadratic.
Hooke's law — The force needed to hold a spring at extension x is proportional to x. The constant of proportionality is the spring constant, and the work to stretch is the integral of that force.
\[ F=kx \;\Longrightarrow\; W=\int_{0}^{d}kx\,dx = \tfrac12kd^{2} \]
The commonest error is to multiply the final force by the distance, which doubles the answer — because the force was smaller than its final value throughout the stretch.
Figure (svg): Hooke's law: force proportional to extension, so work is an integral
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 696-704 — work done by a variable force
Picture it
A linear spring, stretched.
Figure (svg): Hooke's law: force proportional to extension, so work is an integral
The shaded triangle is the work. Its area is half the base times the height, which is why a linear spring's work is half the final force times the distance.
Worked example
Example 6.27. Hooke's law applied.
\[ \text{A spring with } k=100 \text{ N/m is stretched } 0.5 \text{ m from rest. Find the work.} \]
Write the force
Why: Hooke's law.
\[ F = 100 x \]
Identify the representative piece
Why: A short move.
Write its work
Why: Force times distance.
\[ 100 x \,dx \]
Integrate from 0 to 0.5
Why: The power rule.
\[ 50(0.25) \]
State
Why: With units.
\[ 12.5 J \]
Figure (svg): Hooke's law: force proportional to extension, so work is an integral
\[ W = \int_{0}^{0.5}100x\,dx = 12.5 \]
Verify: compare with the wrong product
Why: The final force is 50 newtons, and multiplying by 0.5 metres would give 25 joules — exactly twice the correct answer. The reason is visible in the picture: the force started at zero and reached 50 only at the end, so the average force was 25 newtons, not 50. For a linear spring the work is always half the final force times the distance, which is a useful shortcut and an even more useful check.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 699-701
Fill the middle
Work to stretch a spring from rest.
Fill in the blanks
W = \int_2^___kx\,dx = \tfrac______}kd^___
Why: Integrating a linear force gives a quadratic work with a factor of one half. That half is exactly the difference between the correct answer and the final-force-times-distance error.
Worked example
Checkpoint 6.27. The limits are not from zero.
\[ \text{Find the work to stretch the same spring from } 0.5 \text{ m to } 0.8 \text{ m.} \]
Note the force law is unchanged
Why: Measured from the natural length.
\[ F = 100 x \]
Set the limits
Why: The two extensions.
\[ 0.5\text{ to } 0.8 \]
Integrate
Why: The power rule.
\[ 50(0.64 - 0.25) \]
Evaluate
Why: The difference.
\[ 19.5 J \]
Compare with the first stretch
Why: A shorter move, more work.
\[ 12.5 J\text{ for } a\text{ longer one} \]
Figure (svg): The solution to Worked example stretching further from an already-stretched state shown as a ladder of expressions, one row per legal move
\[ W = \int_{0.5}^{0.8}100x\,dx = 19.5 \]
Verify: explain why a shorter stretch costs more
Why: The first stretch covered half a metre and the second only three tenths, yet the second required more work — because the force is larger throughout the later stretch, running from 50 to 80 newtons rather than from 0 to 50. That is characteristic of a varying force and is why the limits must be measured from the natural length rather than from wherever the stretch begins.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 701-703
Error analysis
A student computes the work to stretch a spring.
Annotate
On: \( W = F\cdot d = (100)(0.5)(0.5) = 25 \text{ J} \)
The elementary formula applies only to a constant force. Whenever a force varies, the product must be replaced by an integral, and the factor of two is the price of forgetting.
Sorting
Does the force change as the object moves?
Sort into buckets
Sort each situation.
The rope is worth noticing: as it is wound up, less of it hangs, so the weight being lifted falls steadily. Any situation where the object being moved changes as the motion proceeds needs an integral.
Two truths and a lie
All three are about springs.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one and it doubles the answer. The force starts at zero and grows linearly, so its average over the stretch is half its final value — which is what the integral computes and what the triangle's area shows.
Prediction
Commit before reasoning.
Predict first
Why is the correct work exactly half the final-force-times-distance answer for a spring?
Correct: Because the force rises linearly from zero.
\[ \text{triangle} = \tfrac12\text{ base}\times\text{height} \]
Why: A linearly increasing quantity averages the mean of its endpoint values, and starting from zero that mean is half the final value. Geometrically the work is the area of a triangle rather than a rectangle of the same height, and a triangle is half its rectangle. For a non-linear force law the factor would be different, so the two is a consequence of Hooke's law rather than of springs as such.
Section
Section 3
Concept
Each thin layer of liquid has a weight depending on the tank's shape at that depth, and must be lifted a distance depending on where it starts. Both vary, and they are different functions.
pumping work — The work to empty a tank is the integral over depth of the layer's weight times the distance it must be lifted. The weight comes from the tank's cross-section and the lift from the geometry of the outlet.
\[ W = \int \rho g A(y)\,(\text{lift}(y))\,dy \]
Setting up a coordinate and stating where its origin is, before writing anything else, is what keeps the two functions apart. Most errors here are sign or reference-level errors.
Figure (svg): Pumping a tank: each layer is lifted a different distance
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 702-710 — work required to pump a tank
Picture it
A tank being emptied over its rim.
Figure (svg): Pumping a tank: each layer is lifted a different distance
The layer near the top barely moves; the one at the bottom travels the tank's whole height. That distance is a separate function of depth from the layer's weight.
Worked example
Example 6.29. Both factors written down.
\[ \text{A cylinder of radius } 2 \text{ m and height } 5 \text{ m, full of water, is pumped over its rim. Find the work.} \]
Set a coordinate
Why: Measure y up from the base.
\[ y\text{ from } 0\text{ to } 5 \]
Write a layer's volume
Why: Cross-section times thickness.
\[ 4 \pi \,dy \]
Write its weight
Why: Density times gravity times volume.
\[ 1000(9.8) (4 \pi) \,dy \]
Write its lift
Why: To the rim at height 5.
\[ 5 - y \]
Integrate
Why: The product, from 0 to 5.
\[ \text{about } 1.54\text{ million joules} \]
Figure (svg): Pumping a tank: each layer is lifted a different distance
\[ W = 39200\pi\int_{0}^{5}(5-y)\,dy = 490000\pi \]
Verify: check against lifting the whole mass from its centre
Why: The tank holds 20 pi cubic metres, about 62.8 tonnes, whose centre of mass is at height 2.5 metres and must reach 5 — a lift of 2.5 metres. That gives 62800 times 9.8 times 2.5, about 1.54 million joules, matching. That shortcut works whenever the lift distance is linear in the depth, and it is a strong check on a long setup.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 705-707
Fill the middle
A layer at height y in a tank of height 5, pumped over the rim.
Fill in the blanks
\text5 = ___ - y
Why: The lift is the destination's height minus the layer's. Using the coordinate itself instead is the section's characteristic error and it survives on a cylinder because the integrand is symmetric.
Worked example
Checkpoint 6.29. Both factors non-constant.
\[ \text{A cone of radius } 2 \text{ and height } 4, \text{ point down, full of water, pumped over the top.} \]
Set a coordinate
Why: Up from the point.
\[ y\text{ from } 0\text{ to } 4 \]
Find the radius at height y
Why: Similar triangles.
\[ r = \frac{y}{2} \]
Write a layer's weight
Why: Density, gravity, area, thickness.
\[ 1000(9.8) \pi(\frac{y}{2}) ^{2} \,dy \]
Write its lift
Why: To the top at height 4.
\[ 4 - y \]
Integrate
Why: Expand and use the power rule.
\[ \text{about } 0.41\text{ million joules} \]
Figure (svg): The solution to Worked example a tank whose cross-section varies shown as a ladder of expressions, one row per legal move
\[ W = 2450\pi\int_{0}^{4}y^{2}(4-y)\,dy = \frac{2450\pi\cdot 64}{3} \]
Verify: compare with a cylinder of the same height
Why: A cylinder of radius 2 and height 4 would hold three times the water and its centre of mass would sit higher, so it would need considerably more work — and indeed it comes to about 1.23 million joules, three times as much. The cone's advantage is that most of its water is near the point, low down but also small in volume. Note that both factors varied here: the cross-section grew with height while the lift shrank, and keeping them apart is the whole difficulty.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 707-709
Trap
\[ W = 39200\pi\int_{0}^{5}y\,dy \]
Use the coordinate as the lift distance
Why: The student integrates against y rather than 5 minus y.
With y measured up from the base, a layer at height y must be lifted 5 minus y to reach the rim, not y.
\[ W = 39200\pi\int_{0}^{5}(5-y)\,dy \]
State where the coordinate's origin is, then write the lift as a distance
Why: The lift is the rim's height minus the layer's.
For a cylinder the two happen to give the same number, since the integrand is symmetric on the interval — which lets the error survive. On a cone or any tapering tank they differ, and the error appears.
Ranking
Pumping a tank.
Put in order
Why: Step a is the one skipped and it is what makes step d unambiguous. Without stating the origin, the lift distance is guesswork, and a sign error there is invisible in the arithmetic that follows.
Sorting
Weight, or lift?
Sort into buckets
Sort each quantity in a pumping problem.
The two groups depend on completely different features: the weight on the tank's shape, the lift on where the outlet is. A tank could be reshaped without moving its outlet, or the outlet raised without reshaping the tank, and only one group would change.
Prediction
Commit before reasoning.
Predict first
Why can a cylinder's pumping work be computed as the total weight times the lift of its centre of mass?
Correct: Because the lift distance is linear in the depth.
\[ \int w(y)(H-y)\,dy = W\left(H-\bar{y}\right) \]
Why: The work is the integral of weight times lift, and when the lift is linear that integral equals the total weight times the lift evaluated at the weighted average position — which is the definition of the centre of mass. The shortcut therefore works for any tank shape as long as the outlet is at a fixed height, and it is a strong independent check on a long setup. Section 6.6 makes the centre of mass its subject.
Section
Section 4
Concept
The pressure on a submerged surface is proportional to the depth. A horizontal strip experiences a nearly uniform pressure, and integrating over the plate's depth gives the total force.
hydrostatic force — The total force a liquid exerts on a submerged vertical surface, equal to the integral of the pressure times the strip's area over the surface's depth.
\[ F = \int \rho g\,(\text{depth})\,w(y)\,dy \]
Because pressure is proportional to depth and not to the amount of liquid, the force on a dam depends on the water's depth and the dam's shape but not on how far back the reservoir extends.
Figure (svg): Hydrostatic force: pressure grows with depth, so strips deeper down push harder
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 708-716 — hydrostatic force and pressure
Picture it
A submerged plate with its strips.
Figure (svg): Hydrostatic force: pressure grows with depth, so strips deeper down push harder
The arrows lengthen with depth because pressure is proportional to it. The lower strips dominate the total, which is why dams are built thicker at the base.
Worked example
Example 6.31. Constant width, varying pressure.
\[ \text{A vertical plate } 3 \text{ m wide extends from } 2 \text{ m to } 5 \text{ m below the surface. Find the force on it.} \]
Set a coordinate
Why: Depth below the surface.
\[ y\text{ from } 2\text{ to } 5 \]
Write the pressure at depth y
Why: Density times gravity times depth.
\[ 1000(9.8) y \]
Write a strip's area
Why: Width times thickness.
\[ 3 \,dy \]
Write its force
Why: Pressure times area.
\[ 29400 y \,dy \]
Integrate from 2 to 5
Why: The power rule.
\[ 14700(25 - 4) \]
Figure (svg): Hydrostatic force: pressure grows with depth, so strips deeper down push harder
\[ F = 29400\int_{2}^{5}y\,dy = 308700 \]
Verify: check against the pressure at the middle depth
Why: The plate's centre is at depth 3.5 metres, where the pressure is 1000 times 9.8 times 3.5, about 34 300 pascals — and the plate's area is 9 square metres, giving about 309 000 newtons. That matches exactly, because the pressure is linear in depth and the plate has constant width, so the average pressure is the pressure at the average depth. For a plate whose width varies the shortcut fails.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 711-713
Fill the middle
Pressure at depth y in a liquid.
Fill in the blanks
p = \rho g\,y
Why: Pressure is proportional to depth and to the liquid's density, and it does not depend on how much liquid is present. That is why the force on a dam depends on the water's depth but not the reservoir's extent.
Worked example
Checkpoint 6.31. Both factors depend on depth.
\[ \text{A triangular plate, } 4 \text{ m wide at the surface and tapering to a point } 3 \text{ m down.} \]
Set a coordinate
Why: Depth below the surface.
\[ y\text{ from } 0\text{ to } 3 \]
Find the width at depth y
Why: Similar triangles.
\[ w = 4(1 - \frac{y}{3}) \]
Write the pressure
Why: Proportional to depth.
\[ 9800 y \]
Write a strip's force
Why: Pressure times area.
\[ 9800 y \times 4(1 - \frac{y}{3}) \,dy \]
Integrate from 0 to 3
Why: Expand and use the power rule.
\[ \text{about } 58 800 N \]
Figure (svg): The solution to Worked example a plate whose width varies shown as a ladder of expressions, one row per legal move
\[ F = 39200\int_{0}^{3}y\left(1-\tfrac{y}{3}\right)dy = 58800 \]
Verify: check why the middle-depth shortcut fails here
Why: The plate's area is 6 square metres and its middle depth is 1.5 metres, which would predict 9800 times 1.5 times 6, about 88 200 newtons — considerably more than the truth. The shortcut fails because the plate is widest where the pressure is least, at the surface, so weighting by area shifts the effective depth upward. Only a constant-width plate lets the average pressure equal the pressure at the average depth.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 713-715
Error analysis
A student computes the force on a submerged plate.
Annotate
On: \( F = pA = 1000(9.8)(5)(9) = 441000 \text{ N} \)
This is the same error as using a spring's final force throughout a stretch. Whenever a factor varies across the object, the elementary product must be replaced by an integral.
Sorting
The middle-depth trick needs a constant width.
Sort into buckets
Sort each plate.
The shortcut is really the statement that force equals pressure at the centroid's depth times the area — and for a constant-width plate the centroid is at the middle depth. For other shapes the centroid is elsewhere, which Section 6.6 computes.
Two truths and a lie
All three are about hydrostatic force.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one and it overestimates, since the pressure is smaller everywhere above the bottom. For the worked example it gives 441 000 newtons against a true 309 000 — an overestimate of about 43 percent.
Prediction
Commit before reasoning.
Predict first
Why does the force on a dam depend on the water's depth but not on how far back the reservoir extends?
Correct: Because pressure depends on depth alone.
\[ p = \rho g h, \quad \text{independent of the volume} \]
Why: The pressure at a point in a liquid is the density times gravity times the depth, with no dependence on how much liquid lies to either side — a consequence of the liquid transmitting pressure equally in all directions. So a dam holding back a small pond of a given depth experiences the same force per unit area as one holding back a lake of the same depth. That is genuinely surprising the first time, and it is why dam design is driven by depth rather than by capacity.
Section
Section 5
Concept
The integrand's units multiplied by the differential's must give the answer's units. A mismatch means the representative piece was described wrongly, and it is visible before any integration.
dimensional analysis — Checking that the units of an expression are consistent with the quantity it claims to compute. In this section it catches almost every setup error.
\[ [\text{integrand}]\times[\text{differential}] = [\text{answer}] \]
Together with a stated coordinate origin, this makes the setups in this section reliable. Both checks take seconds and neither error is visible in the arithmetic afterwards.
Figure (svg): Units as a check on every setup in this section
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 690-716 — setting up physical applications
Picture it
Each integrand times its differential.
Figure (svg): Units as a check on every setup in this section
In every row the units multiply out correctly, which is what confirms the piece was described properly. A row that did not would reveal the error before a single integration.
Worked example
Example 6.32. Dimensional analysis in practice.
\[ \text{Is } W=\int_{0}^{5}1000\cdot 9.8\cdot 4\pi(5-y)\,dy \text{ dimensionally correct?} \]
Check the density's units
Why: Mass per volume.
\[ \text{kg} / m ^{3} \]
Multiply by gravity
Why: Acceleration.
\[ N / m ^{3}, a\text{ weight density} \]
Multiply by the area
Why: Square metres.
\[ \frac{N}{m} \]
Multiply by the lift
Why: Metres.
Multiply by the differential
Why: Metres.
Figure (svg): Units as a check on every setup in this section
\[ \left[\tfrac{\text{kg}}{\text{m}^{3}}\right]\left[\tfrac{\text{m}}{\text{s}^{2}}\right]\left[\text{m}^{2}\right]\left[\text{m}\right]\left[\text{m}\right] = \text{J} \]
Verify: see what a common error would have shown
Why: Omitting the area factor would leave the units as newtons per square metre times metres squared, which is not joules — so the error is visible in the units before any arithmetic. Dimensional analysis will not catch a wrong numerical constant or a sign error, but it catches every missing or extra factor, and those are the commonest mistakes in this section.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 714-715
Matching
Each integral is a product made variable.
Match the pairs
Why: Every row is a formula from elementary physics, and the integral appears only because one factor varies. Recognising which product a problem is about is the first step, and it identifies the integrand immediately.
Worked example
Checkpoint 6.32. One method, five applications.
\[ \text{State the pattern common to every problem in this section.} \]
Identify the elementary formula
Why: For constant factors.
Find which factor varies
Why: With position.
Choose a piece where it is constant
Why: Matching the symmetry.
Write the product for that piece
Why: The elementary formula applies.
Integrate over the object's extent
Why: Chapter 5 routine.
Figure (svg): The solution to Worked example the general pattern shown as a ladder of expressions, one row per legal move
\[ \text{product on a piece} \;\longrightarrow\; \text{integral} \]
Verify: confirm the pattern covers every problem in the section
Why: Mass is density times length, work is force times distance, pumping work is weight times lift, and hydrostatic force is pressure times area — four elementary products, each with one factor varying. Sections 6.1 to 6.4 fit the same description with geometric products. So the whole of Chapter 6 is one method applied to different elementary formulas, which is a more useful thing to remember than five separate setups.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 715-716
Trap
\[ \text{lift} = y \text{ or } 5-y? \quad \text{(the student is unsure)} \]
Write the integral without deciding where y is measured from
Why: The origin was never stated.
The lift distance cannot be written until the coordinate is fixed, and guessing gives a sign error that the arithmetic will not reveal.
\[ \text{let } y \text{ be the height above the base} \;\Longrightarrow\; \text{lift} = 5-y \]
State the coordinate and its origin before anything else
Why: Every subsequent expression depends on it.
Writing one sentence fixing the coordinate is the cheapest error-prevention in this section. It also makes the limits obvious, since they are the coordinate's range over the object.
Fill the middle
A work integral's integrand and differential.
Fill in the blanks
[\textJ]\times[\text___] = ___
Why: Newtons times metres gives joules, confirming the setup computes a work. A mismatch would reveal a missing or extra factor before any integration.
Sorting
Before writing any integral.
Sort into buckets
Sort each item by whether it must be fixed before setting up.
The expected size is worth thinking about even though it is optional, since it catches errors the units cannot — a factor of two, or a sign. Both checks together cover most of what goes wrong.
Prediction
Commit before reasoning.
Predict first
Dimensional analysis catches which kind of error?
Correct: Missing or extra factors.
\[ \text{units catch structure}; \quad \text{estimates catch magnitude} \]
Why: A dropped area or circumference changes the units and shows up immediately, which covers the commonest mistakes in this section. A wrong numerical constant, a sign error or a wrong limit leaves the units intact and passes the check unnoticed. That is why the units check pairs with a size estimate: between them they catch most of what goes wrong, and neither alone is sufficient.
Comparison
Fill the blanks. Each is a product with one varying factor.
Comparison matrix
| Quantity | Elementary product | What varies |
|---|---|---|
| Mass | density times length | the density |
| Spring work | force times distance | the force, with extension |
| Pumping work | weight times lift | both, and differently |
| Hydrostatic force | pressure times area | the pressure, with depth |
The pumping row is the hardest because both factors vary and they are different functions of the same coordinate. Keeping them apart is what a stated coordinate origin is for.
Pattern
Given a physical quantity to compute.
Steps two and five are the two cheap checks. A stated origin makes distances unambiguous, and a units check catches every missing factor — neither error is visible in the arithmetic that follows.
Stewart, Calculus: Early Transcendentals 8e, §6.4 Work §6.4, pp. 455-460
Check
Springs.
Check your understanding
A spring with k = 100 N/m is stretched 0.5 m. What is the work?
Answer: A
Why: The integral of 100x from 0 to 0.5 is half of 100 times 0.25.
Check
Pumping.
Check your understanding
With y measured up from a 5 m tank's base, what is a layer's lift to the rim?
Answer: A
Why: The layer must travel from height y up to height 5.
Check
Hydrostatic force.
Check your understanding
Why does the force on a dam not depend on the reservoir's extent?
Answer: A
Why: The pressure at a point is density times gravity times depth, with no volume dependence.
Real world
A brewery must empty a conical fermentation vessel, point down, through an outlet two metres above its rim, and wants to size the pump. The vessel is four metres tall with a two-metre radius at the top.
Discussion prompt
Explain how the work is computed, why both factors vary, and how the pump's power rating follows.
Hint: Each layer weighs a different amount and travels a different distance.
Answer:
Each thin layer's weight depends on the vessel's radius at that height, which grows linearly from the point; its lift is the distance from that height up to the outlet at six metres above the point. Both vary and they are different functions of the same coordinate — one increasing, the other decreasing.
\[ W = \rho g\pi\int_{0}^{4}\left(\frac{y}{2}\right)^{2}(6-y)\,dy \]
The outlet's height above the rim matters and is easy to forget. Pumping merely to the rim would use 4 minus y; the extra two metres adds a term equal to twice the total weight, which for this vessel is a substantial fraction of the total work.
The pump's rating follows from the time allowed. Work divided by the emptying time gives the average power, and the peak power is higher because the first layers pumped are the ones nearest the top — light, but the deep ones later are heavy AND must be lifted furthest. A pump sized on the average will stall at the end.
Note that this is exactly the pumping setup with a non-trivial lift reference, and the two checks of the last idea both apply: state where the coordinate's origin is, and confirm the units multiply to joules. The units check would catch a dropped area factor; only a size estimate would catch the forgotten two metres.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why is the work to stretch a spring not the final force times the distance?
Correct: Because the force was smaller throughout.
\[ W = \int_{0}^{d}kx\,dx = \tfrac12 kd^{2} = \tfrac12 F_{\text{final}}d \]
Why: The force starts at zero and grows linearly to its final value, so its average over the stretch is half that value — and the work is the area under the force line, a triangle rather than a rectangle. Multiplying by the final force doubles the answer, and the same error appears as using the deepest pressure across a whole submerged plate. Any product with a varying factor must become an integral.
Explain it
They computed a spring's work as the final force times the distance and got twice the right answer.
Discussion prompt
In four sentences or fewer, show them what went wrong.
Hint: Ask what the force was at the start.
Answer:
Ask them what force the spring exerted at the very beginning of the stretch: zero, since it was at its natural length. So it was never 50 newtons except at the last instant, and using 50 throughout assumes a force the spring never had for most of the move.
Sketching force against extension makes it obvious — the work is the area under a rising line, which is a triangle, and a triangle is half its rectangle. That factor of two is exactly what they gained.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For mass, match the piece to the symmetry — a ring for a radial density. For work, replace the product by an integral whenever the force varies. For pumping, state the coordinate's origin and write weight and lift as separate functions. For hydrostatic force, write the width as a function of depth. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the four elementary products of this section side by side, and under each write which factor varies and what the representative piece is. Below, sketch force against extension for a spring, shade the work, and write both the correct answer and the doubled one with a note on where the factor of two comes from. In the middle of the page, draw a tank with one layer marked, label its weight and its lift as two separate arrows, and write the integral with the coordinate's origin stated in words beside it. To the right, draw a submerged plate with pressure arrows lengthening downward and write the force integral. In the lower half, write the four unit chains showing each integrand times its differential giving the right unit. At the bottom, write the five-step procedure with the coordinate statement second.
If your tank drawing has only one arrow on the layer, add the second — the weight and the lift are different quantities pointing in different senses, and drawing them separately is what keeps them from being confused.
Recap
Five things, and all of them are one elementary product made variable.
| If you see | Then |
|---|---|
| A constant density or force | Multiply: no integral needed |
| A density varying with radius | The piece is a ring: include the circumference |
| A spring | Work is half the final force times the distance |
| A tank being pumped | Weight and lift are separate functions of depth |
| An unstated coordinate | Fix it before writing anything |
| A submerged plate | Pressure grows with depth; write the width too |
| Any setup at all | Check the units before integrating |
Section 6.6 closes the physical applications with moments and centres of mass — the point at which an object balances, which is a weighted average computed by exactly this method.
OpenStax Calculus Volume 1, §6.5 Physical Applications §6.5, pp. 594-609 — everything on these slides traces back here
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