CEGEP Calculus II Area, Volume and Arc Length Free · no sign-up

CEGEP Calculus II — Applications of Integration Worksheet

By Marius — private math tutor in Montreal. Updated 30 August 2026

The set where an integral stops being an antidifferentiation exercise and starts measuring a region, a solid or a curve. Area between two curves; volumes of revolution by disks and washers; the same volumes by cylindrical shells; deciding which of the three the geometry actually favours; and arc length, including the engineered radicand that turns a square root into something you can integrate. Almost nothing here is lost in the antiderivative — it is lost in the set-up, which is why every question starts on a sketch. Read them on this page, then print the free PDF when you want room to draw.

Practice worksheet — free PDF

4 pages 6 questions Letter size, print-ready

Download the worksheet Read the questions here

No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 4 harder problems come with the CEGEP Calculus II bundle.

All 6 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

Q1Area Between Two Curves

The curves $y = 2 x - x^{2}$ and $y = x^{2} - 4 x$ enclose a single bounded region.
1. Sketch both curves on the grid, mark the two points where they meet, and shade the enclosed region.
2. Find the area of that region.
A blank Cartesian grid for this question is on the printable PDF.
Q2Volumes by Disks and Washers

The region bounded by $y = x^{2}$ and $y = 2 x$ is revolved about the $x$ -axis. Sketch the region, identify the outer and inner radii of a typical washer, and find the volume of the solid generated.
Q3Volumes by Cylindrical Shells

Let $R$ be the region bounded by $y = x^{2}$ , the $x$ -axis, and the vertical lines $x = 1$ and $x = 3$ . Using the shell method, find the volume of the solid obtained by revolving $R$ about the $y$ -axis. State the radius and the height of a typical shell before integrating.
Q4Choosing Between Disks, Washers and Shells

Let $R$ be the region bounded by $y = x^{2} + 1$ , the $x$ -axis, and the lines $x = 0$ and $x = 2$ . For each axis of revolution below, decide which method the geometry favours, write down the corresponding integral without evaluating it, and say in one or two sentences what makes the other method harder here. A sketch of $R$ will settle both.
1. $R$ revolved about the $y$ -axis.
2. $R$ revolved about the $x$ -axis.
Q5Arc Length

The profile of a skateboard ramp is modelled by $y = \frac{2}{3} x^{3 / 2}$ for $0 \leq x \leq 3$ , with $x$ and $y$ measured in metres. Find the exact length of the ramp surface, then give it to three decimal places.
Q6Synthesis — drawing on several topics in this set

A strip of metal is bent so that its edge follows the curve $y = \frac{x^{3}}{6} + \frac{1}{2 x}$ for $1 \leq x \leq 2$ , with $x$ and $y$ in metres. Let $R$ be the region bounded by this curve, the $x$ -axis, and the lines $x = 1$ and $x = 2$ .
1. Show that $1 + {(\frac{d y}{d x})}^{2}$ is a perfect square, and hence find the exact length of the strip.
2. Find the exact area of $R$ .
3. Find the exact volume of the solid obtained by revolving $R$ about the $x$ -axis.

The 4 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

Which stream is this for? This is CEGEP Calculus II — 201-NYB-05 under the legacy numbering, 201-SN3-RE under the current one — and it sits inside competency 0M03 of the Sciences de la nature programme. The ministerial devis names this material outright: areas of bounded regions, and volumes of solids of revolution by disks and by shells. It also asks for an appropriate graph of the region, which is why the sketch is treated here as part of the answer rather than as scratch work. Arc length is included; surface area of revolution is not part of this course and appears nowhere on the sheet. The integration techniques themselves — parts, trigonometric substitution, partial fractions — have their own worksheet in this course, so every integral below is reachable with substitution, the power rule and the logarithm.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

Area between two curves begins with the sketch, and the devis says so (Q1)

Q1 hands you two parabolas, a grid, and three instructions in order: sketch both curves, mark the two points where they meet, shade the enclosed region — and only then find the area. That order is the method, not a courtesy. The intersections are the limits of integration, and which curve you subtract from which is read off the picture.

A = ∫ (top − bottom) dx or A = ∫ (right − left) dy

The integrand is always the length of one slice, written so that it is non-negative across the whole interval. Solve the two equations simultaneously for the limits, then test a single interior point to settle the order — substituting one convenient x into both expressions is faster and safer than reasoning about which parabola "should" be higher.

Every crossing inside the interval forces a split. "Top minus bottom" is a description of a picture, not a fixed pair of functions: where the curves swap places, the difference changes sign, and a single integral lets the two pieces cancel — a region with obvious area can integrate to zero. Find all the intersection points before you write any limits, split the interval at each one, and subtract in whichever order keeps each integrand positive. A sketch shows you the crossings; the algebra alone hides them.

Disks and washers: square each radius, then subtract (Q2)

Revolve a region about a line and every slice taken perpendicular to that line is a circle or an annulus. A disk when the region reaches the axis along the whole slice; a washer when a gap is left between the region and the axis. Q2 asks you to name the outer and inner radii of a typical washer before integrating, which is exactly where the marks sit.

disk: V = π ∫ R² dx washer: V = π ∫ (R² − r²) dx

π ∫ (R² − r²) dx, never π ∫ (R − r)² dx. The area of an annulus is the large circle minus the small one, πR² − πr²: two areas subtracted, so each radius is squared on its own before anything is taken away. (R − r)² is the square of the gap between the radii, and it is the area of nothing in this picture. Expand the bracket inside the integral and antidifferentiate afterwards.

Which boundary gives the outer radius is a property of the axis, not of the curves. About a horizontal axis a radius is a vertical distance, so the higher boundary is the farther one; about a vertical axis a radius is a horizontal distance, so the boundary further right is the farther one. And a radius is a distance from the axis of revolution, which makes it a subtraction as soon as the axis is a shifted line rather than a coordinate axis — the function value alone is the radius only when the axis is y = 0 or x = 0.

Cylindrical shells: a strip parallel to the axis (Q3)

Take a strip parallel to the axis instead of perpendicular to it, and revolving it sweeps a thin cylindrical tube. Unroll the tube and it is a rectangle: circumference times height times thickness. Q3 asks you to state the radius and the height of a typical shell before integrating, and naming both is what stops the two getting swapped.

shells: V = 2π ∫ (radius)(height) dx radius = distance from the axis to the strip height = top boundary − bottom boundary

Notice what the pairing does. A vertical strip is parallel to a vertical axis, so a vertical axis of revolution and an integral in x go together under the shell method — and that is precisely the pairing disks and washers cannot use, since their slices must be perpendicular to the axis. The radius must also be non-negative for every value in the interval: it is a distance, so an axis that cuts through the region is a signal to stop and think rather than to carry on integrating.

Choosing between disks, washers and shells (Q4)

Q4 is the question the rest of the sheet exists for. One region, two axes of revolution, and for each you must say which method the geometry favours, write the integral without evaluating it, and explain what makes the other method harder. That last part is the real content: both methods are always correct, and usually only one of them is workable.

This is a geometry decision, not a preference — and the wrong choice does not give a wrong answer, it gives a much longer one. Picking the awkward method typically costs two things at once: the boundary curve has to be inverted so it can be written in the other variable, and one solid has to be split into two integrals because the description of the region changes partway along that variable. A region bounded below by an axis and above by a single curve is described by one formula in x and by two in y — so keep x, and let the axis of revolution choose the method rather than the other way round.

Picking the method, in order

Done on a sketch with one representative strip drawn on it. The strip settles everything.

1
Draw the region and the axis of revolution
The line itself goes on the sketch, not just the curves. Shade the region and mark where the boundaries meet — those points give your limits whichever method wins.
2
Choose the variable that describes the region with one formula
Ask which way you can slice so that a typical strip meets the same pair of boundaries all the way across the interval. If slicing the other way makes a boundary change partway, that way needs a split — and that is the cost you are trying to avoid.
3
Compare that strip with the axis
Parallel to the axis means shells. Perpendicular to the axis means disks or washers. This single comparison names the method; you do not pick it, the two decisions above pick it for you.
4
For disks or washers, ask whether the strip touches the axis
Touching along its whole length means no hole, so a disk. A gap between the region and the axis means a washer, and the inner radius is the distance from the axis to the near boundary — often a non-zero constant when the axis is a shifted line.
5
Write the integral, then look at the integrand once
If the method the geometry favoured produced something you cannot antidifferentiate, and the other one inverts cleanly, set both up and take the easier integrand. Inverting a square root gives a polynomial, so the "awkward" method is occasionally the cheap one. Then say why you chose it — that sentence is part of the answer in Q4.

Arc length, and the radicand that has to cooperate (Q5, Q6)

The length of a curve comes from adding up hypotenuses: over a small step the piece of curve is near enough √(dx² + dy²), and factoring dx out of the root gives the formula.

L = ∫ √(1 + (dy/dx)²) dx over [a, b], with dy/dx continuous there

Q5 models a skateboard ramp with a three-halves power, and the point of that exponent is what it does to the radicand: squaring the derivative leaves something a single substitution turns into a power rule. Q6 goes one step further and tells you in advance that 1 + (dy/dx)² is a perfect square — expand it, recognise the square, and the root disappears entirely.

Arc length integrands almost never integrate in closed form, so the ones you are set are built. A curve of the shape ax³ + b/x is engineered so that the two outer terms of (dy/dx)² are squares while the cross term is exactly the −1 that the +1 cancels, leaving a perfect square with a plus sign in the middle. When a question says "show that the radicand is a perfect square", it is telling you the root is meant to be removed algebraically — not estimated, and not left sitting under the integral sign. Check that the bracket is positive on the interval before you drop the root, because √(u²) is |u|.

Two smaller habits. Say that dy/dx is continuous on the interval before applying the formula — for a curve with an x in a denominator that is a real condition, not a ritual. And when a question asks for an exact length and then a decimal, give both: the exact value is the mathematics and the decimal answers the physical question. Sanity-check the result against the straight chord joining the endpoints, which must always come out a little shorter.

The synthesis question: one region, three measurements (Q6)

Q6 puts a curve, the x-axis and two vertical lines around a single region and then asks three different things about it: the length of the curved edge, the area enclosed, and the volume swept when the region is revolved about the x-axis. Three formulas, one picture, and the same derivative and boundary work reused throughout.

Watch how differently the same function enters each integral. Arc length uses the derivative, under a root. Area uses the function itself. Volume uses the function squared — and because the region rests on the axis of revolution, the slices have no hole, so it is a disk integral rather than a washer one. Expand that square before antidifferentiating: a cross term that simplifies to a power of x is what makes the integral tractable, and leaving the bracket unexpanded is what makes it look impossible.

Getting the most out of it

Sketch first, always, and put the axis on the sketch

Competency 0M03 asks for an appropriate graph of the bounded region, so the picture earns marks in its own right — but it also does the work. Which curve is on top, where the two meet, whether the region touches the axis, whether the strip is parallel or perpendicular to it: those questions are answered by one drawing and by nothing else. Sketching is part of the method here, not preliminary to it.

Name the pieces before you write the integral

Radius and height for a shell; outer radius and inner radius for a washer; top and bottom for an area. Write them out as short labelled expressions first. A set-up with the pieces named and one arithmetic slip further down scores far better than a tidy integral that turns out to describe a different solid.

Ask what the other method would have cost

Before committing, spend ten seconds on the alternative: would it need the curve inverted? Would the description of the region change partway, forcing a split? Q4 asks that question outright, and asking it by habit is what makes the choice quick everywhere else. If the two look comparable, set up both — they must agree, which makes the second a free check on the first.

Check the answer against something you can picture

An area should be smaller than the rectangle enclosing it. A solid of revolution should be smaller than the cylinder or cone containing it. An arc should be longer than the chord joining its endpoints. None of these takes more than a few seconds, and each catches the class of error — a dropped factor, a squared difference, a missing π — that the algebra itself will never flag.

Keep exact values exact

Carry fractions, roots, logarithms and π through to the end, and convert to a decimal only when the question asks for one. Rounding partway is how an answer that would have simplified to a clean multiple of π arrives instead as an untidy decimal that no longer checks against anything.

Want the solutions, or something more challenging?

The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the CEGEP Calculus II Solutions Bundle, which is what keeps the rest of the series free.

What else exists for Applications of Integration

Three PDFs · 8 pages · all three are in the bundle below.

Answer key — 3 pages. All 6 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
Challenge problems — 3 pages, 4 problems. A separate sheet at exam-plus difficulty covering the same 5 concepts. Harder than anything on the free sheet.
Challenge answer key — 2 pages. Every challenge problem worked to the same standard, with the checks shown.
PDF, letter size, print-ready.

In the bundle See what's in it Not sold separately

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Common questions

Is this worksheet really free?

Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete CEGEP Calculus II Solutions Bundle, which covers every set at this level.

Which course is this? I have 201-SN3-RE, not 201-NYB-05.

Both are the same course. 201-NYB-05 is the legacy number and 201-SN3-RE the current one, and the content is fixed by ministerial competency 0M03 in the Sciences de la nature programme, so the material is the same across colleges whichever number is on your timetable. This is CEGEP Calculus II, not AP Calculus.

How do I know whether to use disks, washers or shells?

Draw the region, draw the axis of revolution, and draw one representative strip. A strip perpendicular to the axis sweeps a disk or a washer — a disk if it reaches the axis, a washer if a gap is left between the region and the axis. A strip parallel to the axis sweeps a cylindrical shell. Choose the direction of the strip so that it meets the same two boundaries all the way across the region, and the method follows from the geometry rather than from preference.

Why is the washer integrand π(R² − r²) and not π(R − r)²?

Because a washer is a large circle with a small circle removed, and its area is πR² − πr²: two areas subtracted, so each radius is squared before the subtraction. (R − r)² squares the gap between the radii, which is not the area of anything in the picture, and it always comes out smaller. Write the bracket as (outer)² − (inner)² and expand it inside the integral.

Is surface area of a solid of revolution on this worksheet?

No. Surface area of revolution is not part of this course, so it appears nowhere on the sheet. Arc length is, and it is here: the devis names areas of bounded regions and volumes of revolution by disks and by shells; arc length is included alongside them.

Do I really have to draw the region?

Yes, and it is worth marks. The competency asks for an appropriate graph, and the graph is also where the limits, the order of subtraction, the radii and the choice of method all come from. Setting up a volume without a picture is guessing at a solid you have not looked at.

Why does the square root in an arc length problem always disappear?

Because the questions are built that way. A general arc length integrand has no elementary antiderivative, so a curve set as an exercise is chosen to make 1 + (dy/dx)² a perfect square — the cross term in the expansion cancels the −1 and leaves a bracket squared. Expand it, check the bracket is positive on the interval, and the root comes off cleanly.

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Yes. Print and photocopy it for your own classes freely — I just ask that the tutorinmontreal.ca footer stays on the page.

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The same topic at the other level: AP Calculus AB · Applications of Integration.