Integral Calculus 201-SN3-RE (201-NYB) • CEGEP, Montreal

Revision sheet: applications of integration, 201-SN3-RE (201-NYB)

This sheet does not copy out the six formulas of the chapter: they are in your notes and nobody remembers them for long. It sets up the one thing that replaces them all, the SLICE method, then deals with the eight errors that cost marks even when the formula is quoted correctly.

It is written for CEGEP students in the Science program (Sciences de la nature) taking Integral Calculus, 201-SN3-RE (formerly 201-NYB). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

Every application of integration is the same idea: cut into infinitesimal slices, write down what ONE slice is worth, then add up. The formula is not memorized, it is REBUILT from the slice, and that is why you should always draw one.

This chapter is part of 201-SN3-RE (201-NYB), Integral Calculus

The essentials

The slice method, which replaces every formula

• 1. DRAW the region and a representative slice, vertical or horizontal.
• 2. Write down what THIS slice is worth: an area, a volume, a length, a work.
• 3. Identify the variable of integration: the one whose name the thickness of the slice carries, $dx$ or $dy$ .
• 4. Add up, that is, integrate between the extreme values of that variable.
• A slice perpendicular to the axis of rotation gives a DISK or a washer; a parallel slice gives a SHELL.

The same region, two slices: the vertical one, parallel to the

y

-axis, generates a SHELL when you rotate about that axis; the horizontal one, perpendicular, generates a DISK.

Every formula in the chapter can be recovered in thirty seconds this way. Reciting them from memory means risking swapping $r$ and $h$ .

Volumes, according to the direction of the slice

• DISKS: $V=\pi\int_{a}^{b}\left[R(x)\right]^{2}dx$ , when the slice is perpendicular to the axis and the region touches the axis.
• WASHERS: $V=\pi\int_{a}^{b}\left(\left[R\right]^{2}-\left[r\right]^{2}\right)dx$ , when a hole separates the region from the axis.
• SHELLS: $V=2\pi\int_{a}^{b}(\text{radius})(\text{height})\,dx$ , when the slice is parallel to the axis.
• KNOWN CROSS-SECTIONS: $V=\int_{a}^{b}A(x)\,dx$ , where $A(x)$ is the area of the cross-section, square, triangular or semicircular.
• The RADIUS is always the DISTANCE TO THE AXIS, not the value of the function.

Disk: the region touches the axis. Washer: a hole separates it from the axis, and the slice is an annulus. Shell: the slice, parallel to the axis, generates a hollow cylinder.

Washers are a disk minus the inner disk, never $\left(R-r\right)^{2}$ : the square of the difference is not the difference of the squares.

The other applications, one line each

• Arc length: $L=\int_{a}^{b}\sqrt{1+\left[f'(x)\right]^{2}}\,dx$ .
• Area of a surface of revolution about the $x$ -axis: $S=2\pi\int_{a}^{b}f(x)\sqrt{1+\left[f'(x)\right]^{2}}\,dx$ .
• Average value: $\bar{f}=\frac{1}{b-a}\int_{a}^{b}f$ , and the factor $\frac{1}{b-a}$ is not optional.
• Work done by a variable force: $W=\int_{a}^{b}F(x)\,dx$ ; to pump out a liquid, $W=\int\rho g\,A(y)\,d(y)\,dy$ , where $d(y)$ is the distance to be travelled.
• Hydrostatic force: $F=\int\rho g\,h(y)\,w(y)\,dy$ , where $h$ is the DEPTH below the surface and $w$ the width.

The slice of curve is the hypotenuse of a right triangle with legs

dx

and

dy

ds=\sqrt{dx^{2}+dy^{2}}

, hence the square of

f'

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. Taking the radius equal to the function when the axis is elsewhere

the whole question

What not to write

“Rotation about $y=3$ : $V=\pi\int\left[f(x)\right]^{2}dx$ .”

What to write

“The radius is the DISTANCE to the axis: $V=\pi\int\left[3-f(x)\right]^{2}dx$ .”

The axis of rotation is the line

y=3

: the radius of the slice is the SEGMENT between the curve and that axis, that is,

3-\sqrt{x}

, not

\sqrt{x}

Why: The disk formula implicitly assumes an axis through $y=0$ . As soon as the axis moves, the radius becomes a difference, and forgetting it throws the volume off by a factor that is anything but constant. Drawing the slice shows the radius at once.

2. Writing the square of the difference instead of the difference of the squares

the whole question

What not to write

$V=\pi\int\left(R-r\right)^{2}dx$

What to write

$V=\pi\int\left(R^{2}-r^{2}\right)dx$

With

R=3

and

r=1

, the annulus is worth

\pi(9-1)=8\pi

; the disk of radius

R-r=2

is worth only

4\pi

: a washer is not a disk of radius

R-r

Why: A washer is a large disk MINUS a small disk: you subtract AREAS, hence squares. Numerical check with $R=3$ and $r=1$ : the right formula gives $\pi(9-1)=8\pi$ , the wrong one gives $\pi(2)^{2}=4\pi$ , half as much.

3. Mixing up shell and disk

the whole question

What not to write

“Vertical slice, rotation about the $y$ -axis: $V=\pi\int x^{2}\,dy$ .”

What to write

“A vertical slice is PARALLEL to the $y$ -axis: it generates a shell, $V=2\pi\int x\,h(x)\,dx$ .”

Same region, same axis: the slice PARALLEL to the axis generates a shell, the PERPENDICULAR slice generates a disk. The variable follows the thickness.

Why: The criterion is geometric and has no exception: a slice PERPENDICULAR to the axis gives a disk, a PARALLEL slice gives a shell. The landmark that never misleads is the variable of integration, which must be the one of the thickness of the slice.

4. Forgetting to square the derivative in the arc length

the whole question

What not to write

$L=\int_{a}^{b}\sqrt{1+f'(x)}\,dx$

What to write

$L=\int_{a}^{b}\sqrt{1+\left[f'(x)\right]^{2}}\,dx$

Why: The formula comes from Pythagoras on a slice: $dL=\sqrt{dx^{2}+dy^{2}}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx$ . The square comes from the theorem, not from a convention. Without it, a negative derivative can make the radicand negative, which is the warning sign.

5. Forgetting to divide by the length of the interval

the whole question, since the answer is four times too large

What not to write

“The average value of $f$ on $\left[0, 4\right]$ is $\int_{0}^{4}f$ .”

What to write

“ $\bar{f}=\frac{1}{4-0}\int_{0}^{4}f$ .”

\left[0, 4\right]

f(x)=x

takes its values between

0

and

4

: its average

2

is among them (the rectangle has the area of the triangle), the integral

8

is not.

Why: An average is a sum divided by a count; here the count is the LENGTH of the interval. The check is immediate: the average value must lie between the minimum and the maximum of the function, which the integral alone does not.

6. Measuring depth from the bottom

the whole hydrostatic force question

What not to write

“The plate is $y$ metres above the bottom, so the pressure there is $\rho g y$ .”

What to write

“The pressure depends on the depth BELOW THE SURFACE: if the water is $H$ deep, the depth is $H-y$ .”

Pressure is measured from the SURFACE: the slice at height

y

is at depth

H-y

, not

y

Why: The pressure is zero at the surface and greatest at the bottom: a formula that gives zero at the bottom is obviously upside down. Placing the origin and writing explicitly where the surface is, before any computation, settles the question once and for all.

7. Using a formula without drawing the slice

varies, but it is the source of half the errors in the chapter

What not to write

“A volume of revolution, so $V=\pi\int y^{2}\,dx$ , whatever the situation.”

What to write

“I draw the slice, I read its radius and its thickness, then I write its volume. The formula follows.”

Why: The six formulas of the chapter look alike and are easily swapped under pressure. A drawing, on the other hand, does not lie: it gives the radius, the height, the thickness, and therefore the formula, with nothing to memorize.

8. Integrating a shell with respect to the wrong variable

the whole question

What not to write

$V=2\pi\int_{0}^{2}x\,h(x)\,dy$

What to write

$V=2\pi\int_{0}^{4}x\,h(x)\,dx$ , the thickness of the vertical slice being $dx$

Why: The variable of integration is dictated by the THICKNESS of the slice, and the limits are the extreme values of that variable. Mixing up the variable and the limits is the error that gives a plausible but wrong number, hence the hardest one to spot.

Which method to choose

Which method for a volume of revolution

Draw the region, the axis of rotation, then ONE slice. The direction of the slice relative to the axis decides everything.

The axis of rotation is dashed purple: a perpendicular slice touching the axis gives disks; with a hole, washers; a parallel slice, shells; otherwise, the area of the cross-section.

If the slice is PERPENDICULAR to the axis and the region touches the axis → DISKS: $V=\pi\int R^{2}$ , where $R$ is the distance from the curve to the axis

Example: $y=\sqrt{x}$ rotating about the $x$ -axis
If the slice is perpendicular to the axis but a HOLE separates the region from the axis → WASHERS: $V=\pi\int\left(R^{2}-r^{2}\right)$

Example: the region between $y=\sqrt{x}$ and $y=\frac{x}{2}$ rotating about the $x$ -axis
If the slice is PARALLEL to the axis → SHELLS: $V=2\pi\int(\text{radius})(\text{height})$

Example: the same region rotating about the y-axis, in vertical slices
If the problem gives the shape of the cross-sections perpendicular to an axis → KNOWN CROSS-SECTIONS: $V=\int A(x)\,dx$ , writing down the area of the cross-section

Example: square cross-sections of side $f(x)$ give $V=\int\left[f(x)\right]^{2}dx$

The two methods of revolution always give the same volume. When both are possible, choose the one that avoids inverting the function.

Which formula for which physical quantity

Write down what ONE slice is worth, and the formula appears.

If you are looking for the length of a curve → slice of length $\sqrt{dx^{2}+dy^{2}}$ , hence $L=\int\sqrt{1+\left[f'\right]^{2}}\,dx$

Example: a catenary, an arc of a parabola
If you are looking for a work → slice of work $F\,dx$ for a variable force, or $\rho g\,V(y)\,d(y)$ for pumping

Example: compressing a spring, emptying a tank
If you are looking for a pressure force → slice of force $\rho g\,h(y)\,w(y)\,dy$ , with $h$ the depth BELOW the surface

Example: the force on a vertical dam
If you are looking for an average value → $\frac{1}{b-a}\int_{a}^{b}f$ , never the integral alone

Example: the average temperature over a day
If you are looking for a quantity accumulated from a rate → integrate the rate: $Q=\int_{a}^{b}q'(t)\,dt$

Example: a flow rate integrated gives a volume

The universal check in this chapter is dimensional analysis: the product of the units of the slice must give the unit you are looking for.

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.

A volume of revolution, written up for the marker

When to use it: Any question about a volume generated by rotating a region.

1 Draw the region, the axis of rotation, and ONE representative slice.
2 State the method and justify it: “The slice is perpendicular to the axis, so I use washers.”
3 Write the outer radius and the inner radius as DISTANCES to the axis.
4 Set up the complete integral with its limits, which are the extreme values of the thickness variable.
5 Compute and conclude with cubic units.

Concluding sentence

The volume generated is $V=\pi\int_{0}^{4}\left[\left(\sqrt{x}\right)^{2}-\left(\frac{x}{2}\right)^{2}\right]dx=\frac{8\pi}{3}$ cubic units.

The trap: Skipping the drawing of the slice. It is the only step that gives the radius without any memory effort, and the marking scheme often gives it a point of its own.

Marking: Typically 1 point for the sketch and the slice, 1 point for the stated method, 2 points for the integral set up, 1 point for the computation.

Check before you hand in

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

Check the units of the slice

Multiply the units of each factor of the slice and check that you do get a volume, a work or a force, depending on the question.

Shell: $2\pi\times$ radius $\times$ height $\times$ thickness gives length cubed ✓. A product that gives an area signals a missing factor.
Compare with a simple solid

Enclose the solid in a cylinder or a cone whose volume you can compute in your head, and check that the result has the right order of magnitude.

The solid generated by $y=\sqrt{x}$ on $\left[0, 4\right]$ fits inside a cylinder of radius $2$ and height $4$ , of volume $16\pi$ ; the answer $8\pi$ is plausible.
Check an average value by bounding it

The average value always lies between the minimum and the maximum of the function on the interval. An average outside these bounds signals that the factor $\frac{1}{b-a}$ was forgotten.

$f(x)=x$ on $\left[0, 4\right]$ : the average is $2$ , well between $0$ and $4$ ✓; the integral alone is $8$ , outside the bounds.
Redo the volume with the other method

Disks and shells give the same number. When time allows, recomputing by the other route is the surest check.

A volume found by shells and confirmed by washers leaves no doubt about the radius used.
Check that the arc length exceeds the chord

The length of a curve is always greater than or equal to the distance between its endpoints. A smaller result is impossible.

For $y=x^{2}$ from $(0,0)$ to $(1,1)$ , the chord is $\sqrt{2}\approx 1.414$ ; an arc length of $1.479$ is consistent, a length of $1.2$ is not.

The typical problem, taken apart

A volume by washers, as on the exam

The region in the first quadrant is bounded by $y=\sqrt{x}$ and $y=\frac{x}{2}$ .

Compute the volume of the solid generated by rotating it about the $x$ -axis.

The two curves cross at

(0, 0)

and

(4, 2)

; rotating about the

x

-axis, the region leaves a conical HOLE in the middle, hence washers rather than disks.

Step 1

Intersections: $\sqrt{x}=\frac{x}{2}$ gives $x=\frac{x^{2}}{4}$ , that is, $x^{2}-4x=0$ and $x(x-4)=0$ . The limits are $x=0$ and $x=4$ .

Why

As for an area, the limits come from the intersection. Factor rather than divide by $x$ , otherwise the limit $0$ disappears.

Step 2

Test point $x=1$ : $\sqrt{1}=1$ and $\frac{1}{2}=0.5$ . The square root is ABOVE on the whole interval, so it gives the outer radius.

Why

The outer radius is the curve farthest from the axis. A test point identifies it without ambiguity, and reversing it swaps $R$ and $r$ , which gives a negative volume.

Step 3

The vertical slice is PERPENDICULAR to the axis of rotation, and between the two curves the region does not touch the axis: it is a WASHER, with outer radius $R=\sqrt{x}$ and inner radius $r=\frac{x}{2}$ .

Why

Two decisions in one sentence, both read off the drawing: the direction of the slice gives the method, and the hole requires washers rather than disks.

Step 4

$V=\pi\int_{0}^{4}\left[\left(\sqrt{x}\right)^{2}-\left(\frac{x}{2}\right)^{2}\right]dx=\pi\int_{0}^{4}\left(x-\frac{x^{2}}{4}\right)dx$ .

Why

Subtract the SQUARES, that is, areas of disks, and not the radii. This is where the most expensive error of the chapter is made.

Step 5

$\pi\left[\frac{x^{2}}{2}-\frac{x^{3}}{12}\right]_{0}^{4}=\pi\left(8-\frac{64}{12}\right)=\pi\left(8-\frac{16}{3}\right)=\frac{8\pi}{3}$ .

Why

Keep $\pi$ as a factor to the very end: the exact answer is expected, and rounding comes only afterwards, here $8.38$ cubic units.

Step 6

Check: the solid fits inside the cylinder of radius $2$ and length $4$ , of volume $16\pi$ . The answer $\frac{8\pi}{3}\approx 2.67\pi$ is one sixth of it, which is consistent with how thin the region is ✓.

Why

Bounding by a simple solid catches errors by a factor of $2$ or $\pi$ , the ones that an algebraic reread lets through.

The conclusion, written out

The volume generated is $\frac{8\pi}{3}\approx 8.38$ cubic units, computed by washers with $R=\sqrt{x}$ and $r=\frac{x}{2}$ on $\left[0, 4\right]$ .

The classic mistake on this problem: Writing $\pi\int\left(\sqrt{x}-\frac{x}{2}\right)^{2}dx$ . This integral is worth $\frac{8\pi}{15}$ , five times less: a washer is a difference of AREAS, never the square of a difference of radii.

Learn by heart

• One single method: draw ONE slice, write down what it is worth, add up.
• Slice PERPENDICULAR to the axis: disks or washers. PARALLEL slice: shells.
• The radius is the DISTANCE TO THE AXIS, not the value of the function.
• Washers: $\pi\int\left(R^{2}-r^{2}\right)$ , never $\pi\int\left(R-r\right)^{2}$ .
• Shells: $2\pi\int(\text{radius})(\text{height})$ , integrated along the thickness of the slice.
• Arc length: $\int\sqrt{1+\left[f'\right]^{2}}$ ; the square comes from Pythagoras.
• Average value: $\frac{1}{b-a}\int f$ , between the minimum and the maximum.
• Hydrostatics: depth is measured BELOW THE SURFACE, never from the bottom.

Frequently asked questions

When should I use disks and when should I use shells?

Look at the slice relative to the axis of rotation. A slice perpendicular to the axis generates a disk or a washer, a parallel slice generates a cylindrical shell. Both methods always give the same volume, so choose the one that avoids inverting the function.

How do I find the radius when the axis of rotation is not a coordinate axis?

The radius is the distance between the curve and the axis, so it is a difference. For a rotation about the line y equals three, the radius is three minus the value of the function. Drawing the slice and measuring the segment that joins it to the axis gives the answer without any memory effort.

Why do we subtract the squares in the washer formula?

Because a washer is a large disk from which a small disk has been removed. So you subtract areas, which are proportional to the squares of the radii. The square of the difference of the radii gives a much smaller number with no geometric meaning.

Where does the square root in the arc length formula come from?

From the Pythagorean theorem applied to an infinitesimal slice. A small piece of curve is the hypotenuse of a triangle with sides d x and d y, so its length is the square root of the sum of the squares. Factoring out d x gives the square root of one plus the square of the derivative.

How do I compute the average value of a function?

Integrate the function over the interval, then divide by the length of that interval. The division is not optional: it is what turns a sum into an average. The immediate check is that the value obtained must lie between the minimum and the maximum of the function.

Practise it

Corrected exercises: Applications of integration, 201-SN3-RE (201-NYB)

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.

15 corrected exercises
150 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-applications-of-integration. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: applications of integration, 201-SN3-RE (201-NYB)

The essentials

The slice method, which replaces every formula

Volumes, according to the direction of the slice

The other applications, one line each

The mistakes that cost marks

1. Taking the radius equal to the function when the axis is elsewhere

2. Writing the square of the difference instead of the difference of the squares

3. Mixing up shell and disk

4. Forgetting to square the derivative in the arc length

5. Forgetting to divide by the length of the interval

6. Measuring depth from the bottom

7. Using a formula without drawing the slice

8. Integrating a shell with respect to the wrong variable

Which method to choose

Which method for a volume of revolution

Which formula for which physical quantity

How the answer is expected to be written

A volume of revolution, written up for the marker

Check before you hand in

The typical problem, taken apart

A volume by washers, as on the exam

Learn by heart

Frequently asked questions

When should I use disks and when should I use shells?

How do I find the radius when the axis of rotation is not a coordinate axis?

Why do we subtract the squares in the washer formula?

Where does the square root in the arc length formula come from?

How do I compute the average value of a function?

Corrected exercises: Applications of integration, 201-SN3-RE (201-NYB)

See also

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