MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: work (MATH 141)

Q: How do I set up the integral for pumping water out of a tank?

Slice the water into horizontal layers. For the layer at height y, write its area from the shape of the tank, its weight as 9800 newtons per cubic metre times area times thickness, and the distance from that layer to the outlet. Multiply weight by distance and integrate over the heights where the water actually is, not over the whole tank.

Q: Why is the work to stretch a spring not force times distance?

Because the force of a spring is not constant: it is zero at the natural length and grows in proportion to the stretch. Force times distance uses the final force for the whole stretch and gives exactly twice the right answer. The work is the area of the triangle under the force line, one half k times the stretch squared.

Q: Do I multiply by g when the tank problem is in pounds and feet?

No. The pound is a unit of force, so 62.5 pounds per cubic foot is already the weight of a cubic foot of water. Multiply it by the volume of the layer and by the distance, and the answer comes out in foot-pounds. Only in SI, where 1000 kilograms per cubic metre is a mass, do you multiply by g to get 9800 newtons per cubic metre.

Q: How do I find the work to lift a hanging cable or chain?

Slice the cable, not the path. The piece at depth x below the top weighs the linear density times its length and rises x, so the work is the integral of density times x from zero to the length. For a uniform cable this is the total weight times half the length. A load at the end is added separately, its weight times the full length.

Q: Is half a tank half the work to pump out?

No, it is more than half. The water left in a half-full tank is the bottom half, the part farthest from the outlet. For a vertical cylinder pumped over the top it is three quarters of the work of the full tank, and for a trough with a triangular section, half the depth holds a quarter of the water but still costs half the work.

This sheet is not a summary of section 6.4 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on work problems in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every number below is done by hand, as on the exam: $\rho g = 9800$ N/m $^3$ in SI, $62.5$ lb/ft $^3$ in imperial, and every tank answer is an exact multiple of $\pi$ .

The thread of the chapter

Work is force times distance only on a slice where BOTH are constant, so the first question is always what varies. A force that changes along the path: slice the PATH, $W = \int F(x)\,dx$ . Pieces that travel different distances: slice the OBJECT, and give each slice its own weight and its own distance, from where it IS to where it is DELIVERED.

This chapter is part of MATH 141, Calculus 2 (McGill)

The essentials

Slice what varies

• $W = Fd$ holds only for a CONSTANT force along the motion. Units: $1$ J $= 1$ N $\cdot$ m; in imperial, ft-lb.
• The FORCE changes along the path (spring, leaking bucket, gravity): slice the path, $W = \int_a^b F(x)\,dx$ , the signed area under the force graph.
• The DISTANCE changes from piece to piece (cable, chain, water in a tank): slice the object, $W = \int (\text{weight of a slice}) \times (\text{its distance})$ .
• Tank: a layer of area $A(y)$ and thickness $\Delta y$ weighs $\rho g\,A(y)\,\Delta y$ , with $\rho g = 9800$ N/m $^3$ , or $62.5\,A(y)\,\Delta y$ lb in imperial.
• The distance $D(y)$ goes from where the layer IS to where it is DELIVERED (the rim, an outlet above it). It is $y$ only if the axis was chosen that way.

Left, the force varies: a strip of the path costs

F(x)\,\Delta x

. Right, the distance varies: a layer weighs

\rho g\,A(y)\,\Delta y

and travels

D(y)

to the outlet, not

y

Before any integral, write one line: what is sliced, what one slice weighs (or what the force is on it), and how far it goes. That line is the method mark, and it decides every bound that follows.

The five families, and what each one gives you for free

• Spring: find $k$ from one data pair, $F = kx$ with $x$ a STRETCH; then $\int_{x_1}^{x_2} kx\,dx = \frac{k}{2}(x_2^2 - x_1^2)$ .
• Hanging cable of $\delta$ N/m, length $L$ : the piece at depth $x$ rises $x$ , so $W = \int_0^L \delta x\,dx = \frac{\delta L^2}{2}$ . A load at the end adds its weight times $L$ , no integral.
• Tank: $W = \int_a^b \rho g\,A(y)\,D(y)\,dy$ , with $a$ and $b$ the heights where the liquid IS.
• Leaking bucket: weight as a function of HEIGHT, through the time: at speed $v$ , height $y$ is reached at $t = \frac{y}{v}$ .
• Gravitation: $F(r) = \frac{mgR^2}{r^2}$ , $r$ from the centre of the Earth; $W = mgR^2\left(\frac{1}{R} - \frac{1}{R + h}\right) < mgh$ .

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

One slice, one product: the rules that exist and the two that do not

Read a line as: in this situation, cut this way, and one slice costs this much work. The examples are all done by hand. The red lines are the shortcuts students write, each with the computation that kills it.

Situation	What you slice	Work of one slice
force $F(x)$ along $x$	the path	$F(x)\,\Delta x$
Example: $F = 3x^2 + 2x$ from $1$ to $3$ : $\left[x^3 + x^2\right]_1^3 = 34$ J.
spring, $F = kx$	the path (stretches)	$kx\,\Delta x$
Example: $k = 200$ , stretch $0.2$ to $0.3$ : $100(0.09 - 0.04) = 5$ J.
hanging cable, $\delta$ N/m	the cable, piece at depth $x$	$\delta\,\Delta x \cdot x$
Example: $20$ m at $6$ N/m: $\int_0^{20} 6x\,dx = 1200$ J.
tank	the liquid, layer at height $y$	$\rho g\,A(y)\,\Delta y \cdot D(y)$
Example: cylinder, radius $2$ , height $5$ , over the top: $9800 \cdot 4\pi\int_0^5 (5 - y)\,dy = 490\,000\pi$ J.
leaking bucket	the path	$w(y)\,\Delta y$
Example: $180 - 4y$ N over $30$ m: $\int_0^{30} (180 - 4y)\,dy = 3600$ J.
spring	nothing	final force $\times$ stretch no such rule
Example: $40 \times 0.2 = 8$ J, but $\int_0^{0.2} 200x\,dx = 4$ J. What to do: Integrate $kx$ between the two stretches; the triangle is half the rectangle.
tank	nothing	total weight $\times$ height no such rule
Example: cylinder above: $196\,000\pi \times 5 = 980\,000\pi$ J, twice the true $490\,000\pi$ J. What to do: Slice into layers; each layer has its own distance $D(y)$ .

Every blue line is a Riemann sum waiting to become an integral: one slice, one product, then $\int$ . A red line multiplies two numbers that were never constant at the same time.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Multiplying the final force of a spring by the stretch

the whole question: the integral it skips is the chapter

What not to write

“ $40$ N holds it $0.2$ m beyond natural length, so $W = 40 \times 0.2 = 8$ J.”

What to write

“ $k = \frac{40}{0.2} = 200$ N/m and $W = \int_0^{0.2} 200x\,dx = 100(0.2)^2 = 4$ J.”

The shaded triangle under

F = 200x

is the work,

4

J. The dashed rectangle is what

40 \times 0.2

computes: it charges

40

N from the first millimetre.

Why: The spring resists with $0$ N at the start and $40$ N only at the end. The work is the triangle under $F = kx$ , half the rectangle that $Fd$ computes.

2. Taking the length of the spring for its stretch

2 marks, and every later part that reuses the bounds

What not to write

“Natural length $20$ cm, stretched from $30$ to $40$ cm: $W = \int_{0.3}^{0.4} 100x\,dx = 3.5$ J.”

What to write

“The stretches are $0.1$ and $0.2$ m: $W = \int_{0.1}^{0.2} 100x\,dx = 50(0.04 - 0.01) = 1.5$ J.”

Why: In $F = kx$ , $x$ is measured from the NATURAL length, not from the wall. Subtract the natural length from both lengths before writing the bounds.

3. Lifting every piece of a cable the full length

the whole question

What not to write

“The $20$ m cable at $6$ N/m weighs $120$ N and goes up $20$ m: $W = 2400$ J.”

What to write

“The piece at depth $x$ rises $x$ : $W = \int_0^{20} 6x\,dx = 1200$ J.”

Why: Only the bottom piece rises $20$ m; the top one barely moves. For a uniform cable the average rise is half the length, which is why the right answer is exactly half.

4. Writing the distance of a layer as y by reflex

the whole question, and a factor of three here

What not to write

“Cone vertex down, $6$ m high, radius $3$ m, full: $W = \int_0^6 2450\pi y^2 \cdot y\,dy = 793\,800\pi$ J.”

What to write

“The layer at height $y$ goes up to the rim at $6$ : $W = \int_0^6 2450\pi y^2(6 - y)\,dy = 264\,600\pi$ J.”

Why: $y$ says where the layer IS; the distance is from there to the outlet. On a cylinder the two integrals happen to agree, which is why the habit survives until the first cone.

5. Integrating over the empty part of the tank

2 marks

What not to write

“Cylinder, radius $2$ , height $5$ , water $3$ m deep: $W = 39\,200\pi\int_2^5 (5 - y)\,dy$ .”

What to write

“The water is between $0$ and $3$ : $W = 39\,200\pi\int_0^3 (5 - y)\,dy = 411\,600\pi$ J.”

Why: The bounds are where the LIQUID is, the destination is in $D(y)$ . Mixing the two, or writing $D(y) = 3 - y$ (up to the water surface instead of the rim), are the two faces of the same slip.

6. Reading the similar triangle upside down

the whole integral, by a factor of sixteen

What not to write

“Cone $6$ m high, rim radius $3$ m, vertex down: $\frac{r}{y} = \frac{6}{3}$ , so $r = 2y$ .”

What to write

“ $\frac{r}{y} = \frac{3}{6}$ , so $r = \frac{y}{2}$ : at the rim, $y = 6$ gives $r = 3$ .”

Why: Test the formula at the rim before using it: it must return the rim radius. Vertex UP, the radius shrinks as $y$ grows, $r = \frac{6 - y}{2}$ .

7. Multiplying pounds by g

1 to 2 marks, and an answer 32 times too large

What not to write

“A layer of volume $V$ ft $^3$ weighs $62.5 \times 32 \times V$ lb.”

What to write

“It weighs $62.5V$ lb: the pound is a force, and $62.5$ lb/ft $^3$ is already a weight density.”

Why: In SI, $1000$ kg/m $^3$ is a MASS density and needs $g$ : $\rho g = 9800$ N/m $^3$ . In imperial the data is already a weight. Read the unit before deciding whether $g$ enters.

8. Checking a cone with total weight times average distance

2 marks, and a false check that confirms a wrong answer

What not to write

“Cone vertex down, full: weight $176\,400\pi$ N, average distance $3$ m, so $W = 529\,200\pi$ J.”

What to write

“The layers do not weigh the same, so the shortcut fails: $W = 264\,600\pi$ J, an effective distance of $1.5$ m.”

Same cone, same volume

\frac{16\pi}{3}

^3

of water: vertex down the wide layers are near the outlet, vertex up they are far from it, and the work triples.

Why: Total weight times average distance is legitimate only when every layer weighs the same, a cylinder or a uniform cable. In a cone the heavy layers sit where the distance is short or long, and the integral weighs them.

9. Believing that half full means half the work

1 to 2 marks

What not to write

“The cylinder is half full, so pumping it costs $\frac{1}{2} \times 490\,000\pi$ J.”

What to write

“ $39\,200\pi\int_0^{2.5} (5 - y)\,dy = 367\,500\pi$ J, three quarters of the full tank.”

Why: The water left in a half-full tank is the BOTTOM half, the half that travels farther. For the triangular trough of the set, half the depth is a quarter of the water and half the work.

Which method to choose

What to slice, decided by the FORM of the statement

Read what the statement gives you before writing any integral

If a force is given as a function of position, or read on a graph → slice the path: $W = \int_a^b F(x)\,dx$ , signed area

Example: graph above then below the axis: add the triangles with their signs
If a spring, with a force or a work at some length → convert lengths to STRETCHES, find $k$ , integrate $kx$

Example: natural $20$ cm, from $30$ to $40$ cm: bounds $0.1$ and $0.2$
If a cable or chain hanging or being lifted → slice the cable: piece at depth $x$ rises $x$ ; a load at the end is weight times length

Example: $\int_0^{20} 6x\,dx + 100 \times 20 = 3200$ J
If a liquid pumped out of a tank → slice the liquid in layers: $\rho g\,A(y)\,D(y)$ , bounds where the liquid is

Example: $A(y)$ : constant (cylinder), similar triangles (cone, trough), Pythagoras (sphere)
If a weight that changes during the lift (leak, sand, fuel) → write the weight as a function of HEIGHT through the speed, then slice the path

Example: $0.5$ m/s and $2$ N/s: $180 - 4y$ newtons
If a lift over a distance comparable to the Earth's radius → $F = \frac{mgR^2}{r^2}$ , integrate in $r$ from the CENTRE

Example: from $R$ to $2R$ : $\frac{mgR}{2}$ , half of $mgh$

Two families can meet in one problem: a bucket on a rope is a constant weight, a varying weight and a cable at once. Write each piece separately, with its own method, and add at the end.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Setting up a pumping problem

When to use it: Any question that asks for the work to pump, empty or lift a liquid out of a container

1 Draw the tank, place the origin and say which way $y$ points, in words: y upward from the bottom, or y downward from the rim.
2 Write the area $A(y)$ of the layer at $y$ from the geometry, and test it at one end (the rim must give the rim radius).
3 Write the weight of the layer: $\rho g\,A(y)\,\Delta y$ with $\rho g = 9800$ N/m $^3$ , or $62.5\,A(y)\,\Delta y$ lb.
4 Write the distance $D(y)$ from the layer to the outlet, not to the water surface.
5 Write the integral with bounds where the liquid IS, evaluate it exactly, and give the unit.

Concluding sentence

“With $y$ upward from the bottom, the layer at height $y$ has volume $4\pi\,\Delta y$ , weighs $39\,200\pi\,\Delta y$ N and rises $5 - y$ m, so $W = \int_0^5 39\,200\pi(5 - y)\,dy = 490\,000\pi$ J.”

The trap: Skipping the sentence and writing the integral directly: when the bounds or the distance are wrong, the marker cannot give the method marks, because there is no method on the page.

Marking: Typically 1 mark for the area of the layer, 1 for the weight, 1 for the distance, 1 for the bounds and 1 for the evaluation: four of the five are earned before integrating.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Bracket the answer

The work lies between the total weight times the smallest distance and the total weight times the largest one. An answer outside is wrong, whatever the integral says.

Cylinder above, full: weight $196\,000\pi$ N, distances from $0$ to $5$ m, and $490\,000\pi$ lies between $0$ and $980\,000\pi$ .
Raise the outlet by one unit

Moving the outlet up by 1 m adds exactly the total weight times 1 m. Recomputing the integral both ways checks the distance function.

Outlet $1$ m above the cylinder: $686\,000\pi - 490\,000\pi = 196\,000\pi$ , the weight of the water.
Test the radius at the rim

Any formula for the area of a layer must give the known dimension at the top or the bottom. One substitution, three seconds.

$r = \frac{y}{2}$ at $y = 6$ gives $3$ , the rim radius; $r = 2y$ would give $12$ .
Read the unit

Newtons times metres give joules, pounds times feet give foot-pounds. A g next to a weight in pounds, or a missing g next to kilograms, shows up here.

$1250y\,\Delta y$ lb times $(2 - y)$ ft: foot-pounds, with no $32$ anywhere.

The typical problem, taken apart

A conical tank, partly full, pumped to an outlet above the rim

A tank has the shape of a right circular cone, vertex down, $4$ m high and $2$ m in radius at the top. It contains water $2$ m deep. Compute the work to pump all the water to an outlet $1$ m above the rim. Use $\rho g = 9800$ N/m $^3$ .

No calculator: give the exact value.

Step 1

Origin at the vertex, $y$ upward. The water occupies $0 \le y \le 2$ ; the rim is at $4$ , the outlet at $5$ .

Why

Three heights written before any formula: the bounds come from the water, the distance from the outlet. Mixing them is the most frequent error.

Step 2

Similar triangles: $\frac{r}{y} = \frac{2}{4}$ , so $r = \frac{y}{2}$ (at the rim, $y = 4$ gives $2$ ). Area $A(y) = \frac{\pi y^2}{4}$ .

Why

The test at the rim takes three seconds and catches the ratio read upside down, which would multiply the answer by sixteen.

Step 3

Weight of the layer: $9800 \cdot \frac{\pi y^2}{4}\,\Delta y = 2450\pi y^2\,\Delta y$ N. Distance to the outlet: $5 - y$ .

Why

The distance goes to the OUTLET, not to the rim and not to the water surface at $2$ .

Step 4

$W = \int_0^2 2450\pi y^2(5 - y)\,dy = 2450\pi\left[\frac{5y^3}{3} - \frac{y^4}{4}\right]_0^2 = 2450\pi\left(\frac{40}{3} - 4\right) = 2450\pi \cdot \frac{28}{3} = \frac{68\,600\pi}{3}$ J.

Why

Expand $y^2(5 - y) = 5y^2 - y^3$ before integrating; each term is a power rule, and the exact fraction is the expected answer.

Step 5

Bracket: the water has volume $\int_0^2 \frac{\pi y^2}{4}\,dy = \frac{2\pi}{3}$ , weight $\frac{19\,600\pi}{3}$ N, and every layer travels between $3$ and $5$ m. So $\frac{58\,800\pi}{3} \le W \le \frac{98\,000\pi}{3}$ , and $\frac{68\,600\pi}{3}$ is inside.

Why

The check of the verifications block, done in one line. It would catch a distance written $y$ or a wrong outlet height.

The conclusion, written out

“The work to pump the water to the outlet is $W = \frac{68\,600\pi}{3}$ J, about $7.2 \times 10^4$ J.”

The classic mistake on this problem: Writing the distance $4 - y$ (to the rim, forgetting the outlet), which gives $\frac{49\,000\pi}{3}$ J, or integrating from $0$ to $4$ as if the tank were full.

Learn by heart

• $W = Fd$ only for a constant force; otherwise $W = \int_a^b F(x)\,dx$ .
• Hooke: $F = kx$ , $x$ a stretch from the NATURAL length; $W = \frac{k}{2}(x_2^2 - x_1^2)$ .
• Cable of $\delta$ N/m and length $L$ wound up: $\frac{\delta L^2}{2}$ , never $\delta L \times L$ .
• Tank: $W = \int_a^b \rho g\,A(y)\,D(y)\,dy$ , bounds where the liquid is, $D(y)$ to the outlet.
• $\rho g = 9800$ N/m $^3$ in SI; $62.5$ lb/ft $^3$ in imperial, with no $g$ .
• Cone: similar triangles, tested at the rim. Sphere: Pythagoras, origin at the centre.
• Half full is never half the work: the bottom water travels farthest.
• Gravity: $F = \frac{mgR^2}{r^2}$ , $r$ from the centre; the escape work is $mgR$ .

Frequently asked questions

How do I set up the integral for pumping water out of a tank?

Slice the water into horizontal layers. For the layer at height y, write its area from the shape of the tank, its weight as 9800 newtons per cubic metre times area times thickness, and the distance from that layer to the outlet. Multiply weight by distance and integrate over the heights where the water actually is, not over the whole tank.

Why is the work to stretch a spring not force times distance?

Because the force of a spring is not constant: it is zero at the natural length and grows in proportion to the stretch. Force times distance uses the final force for the whole stretch and gives exactly twice the right answer. The work is the area of the triangle under the force line, one half k times the stretch squared.

Do I multiply by g when the tank problem is in pounds and feet?

No. The pound is a unit of force, so 62.5 pounds per cubic foot is already the weight of a cubic foot of water. Multiply it by the volume of the layer and by the distance, and the answer comes out in foot-pounds. Only in SI, where 1000 kilograms per cubic metre is a mass, do you multiply by g to get 9800 newtons per cubic metre.

How do I find the work to lift a hanging cable or chain?

Slice the cable, not the path. The piece at depth x below the top weighs the linear density times its length and rises x, so the work is the integral of density times x from zero to the length. For a uniform cable this is the total weight times half the length. A load at the end is added separately, its weight times the full length.

Is half a tank half the work to pump out?

No, it is more than half. The water left in a half-full tank is the bottom half, the part farthest from the outlet. For a vertical cylinder pumped over the top it is three quarters of the work of the full tank, and for a trough with a triangular section, half the depth holds a quarter of the water but still costs half the work.

Practise it

Corrected exercises: Work, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Volumes: cylindrical shells Next sheet Arc length and surface area

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Get in touch for a first session. Work problems are where the integral stops being a technique and becomes a model: slice, weigh, measure a distance, add.

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Revision sheet: work (MATH 141)

The essentials

Slice what varies

The five families, and what each one gives you for free

The rules in table form

One slice, one product: the rules that exist and the two that do not

The mistakes that cost marks

1. Multiplying the final force of a spring by the stretch

2. Taking the length of the spring for its stretch

3. Lifting every piece of a cable the full length

4. Writing the distance of a layer as y by reflex

5. Integrating over the empty part of the tank

6. Reading the similar triangle upside down

7. Multiplying pounds by g

8. Checking a cone with total weight times average distance

9. Believing that half full means half the work

Which method to choose

What to slice, decided by the FORM of the statement

How the answer is expected to be written

Setting up a pumping problem

Check before you hand in

The typical problem, taken apart

A conical tank, partly full, pumped to an outlet above the rim

Learn by heart

Frequently asked questions

How do I set up the integral for pumping water out of a tank?

Why is the work to stretch a spring not force times distance?

Do I multiply by g when the tank problem is in pounds and feet?

How do I find the work to lift a hanging cable or chain?

Is half a tank half the work to pump out?

Corrected exercises: Work, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?