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Applications of Integrals

Slice it thin, add it up.

An integral adds up a huge number of thin pieces. Choose the right pieces and the same idea measures the area trapped between two curves, the volume of a vase or a sphere, the work needed to stretch a spring or pump out a tank, and even the length of a curve.

On this page: area between curves and volumes of revolution with graphs you can explore, shells, average value, work, a preview of arc length, six worked examples, nine practice problems and an eight-question quiz.

Explore the area Jump to practice

The region between y = x and y = x² is a stack of thin vertical strips. Each strip is top − bottom tall, and adding them gives an area of exactly 1/6.

Section IArea between curves

Cut the region into thin vertical strips. A strip at x has height top − bottom and width dx. If f(x) ≥ g(x) on a ≤ x ≤ b, adding the strips gives

Area = ∫_a^b (f(x) − g(x)) dx

The recipe: (1) sketch both curves; (2) solve f(x) = g(x) to find where they meet, which often gives a and b; (3) decide which curve is on top; (4) integrate top minus bottom. If the curves cross inside the interval, split it at the crossing so every piece is top minus bottom, otherwise areas cancel. For regions that are easier to slice sideways, integrate right − left with respect to y.

This explorer needs JavaScript turned on. Here is what it shows. y = x and y = x² meet at 0 and 1, enclosing area 1/6. y = 4 − x² and y = x + 2 meet at −2 and 1, enclosing area 9/2. y = √x and y = x/2 meet at 0 and 4, enclosing area 4/3. y = sin x and y = cos x meet at π/4 and 5π/4, with sin on top, enclosing area 2√2 ≈ 2.828427.

Section IIVolumes of revolution: disks and washers

Spin a region about the x-axis and it sweeps out a solid. Slice the solid perpendicular to the axis: each slice is a thin disk of radius R(x) and thickness dx, with volume about πR(x)² dx. Adding the slices,

V = π ∫_a^b R(x)² dx

If the region does not touch the axis, each slice is a washer, a disk with a hole. Its outer radius R comes from the curve farther from the axis, and its inner radius r from the nearer one:

V = π ∫_a^b (R(x)² − r(x)²) dx

Square each radius separately: R² − r² is not (R − r)².

This explorer needs JavaScript turned on. Here is what it shows. Spinning the region under y = √x for 0 ≤ x ≤ 4 gives volume π∫x dx = 8π ≈ 25.132741. Spinning the half-disk under y = √(4 − x²) gives a sphere of radius 2 with volume 32π/3 ≈ 33.510322. Spinning the region between y = x and y = x² for 0 ≤ x ≤ 1 gives washers with R = x and r = x², and volume 2π/15 ≈ 0.418879.

Section IIICylindrical shells, briefly

Spinning about the y-axis instead? Keep the vertical strips. A strip at x sweeps out a thin cylindrical shell of radius x, height f(x) and thickness dx. Unrolled, it is nearly a flat sheet of area 2πx · f(x), so

V = 2π ∫_a^b x f(x) dx

Shells shine when washers would force you to solve for x in terms of y. Both methods always give the same volume; choose whichever integral is easier.

Section IVAverage value, revisited

As on the Fundamental Theorem page, the average value of f on a ≤ x ≤ b is the total divided by the width:

f_avg = 1b − a ∫_a^b f(x) dx

For example, the average of √x on 0 ≤ x ≤ 4 is ¼ · 16/3 = 4/3: a rectangle 4 wide and 4/3 tall has the same area as the region under the curve.

Section VWork

For a constant force, work is force × distance. When the force changes, slice the motion into tiny steps and add: W = ∫_a^b F(x) dx

Springs. Hooke’s law says a spring stretched x metres past its natural length pulls back with force F = kx. Stretching it from 0 to d takes ∫₀^d kx dx = ½kd² joules.

Pumping. Here the distance varies: slice the liquid into thin horizontal layers, find each layer’s weight (density × g × volume) and how far it must be lifted, multiply, and integrate over the depth. Water weighs about 9800 N per cubic metre (density 1000 kg/m³ times g ≈ 9.8 m/s²). Example 6 works one through.

Section VIOptional preview: arc length

Zoom in on a smooth curve and a tiny piece is almost a straight segment with sides dx and dy. By Pythagoras its length is √(dx² + dy²) = √(1 + f′(x)²) dx, so

L = ∫_a^b √(1 + f′(x)²) dx

For y = ⅔x^3/2 on 0 ≤ x ≤ 3, f′(x)² = x, so L = ∫₀³ √(1 + x) dx = ⅔(8 − 1) = 14/3. Most curves give integrals with no simple formula, which is why arc length is usually met again later with numerical methods.

Section VIIWorked examples

Example 1: area between a parabola and a line

Find the area between y = 4 − x² and y = x + 2.

4 − x² = x + 2, so x² + x − 2 = 0, x = −2 or 1IntersectionsAt x = 0: 4 > 2, so the parabola is on topTest a point∫₋₂¹ (2 − x − x²) dxTop minus bottom[2x − x²/2 − x³/3] = 7/6 − (−10/3) = 9/2Answer

✓ A midpoint sum with n = 1000 gives 4.500002.

Example 2: disks

Spin the region under y = √x, 0 ≤ x ≤ 4, about the x-axis.

R(x) = √x, so R² = xDisk radiusπ ∫₀⁴ x dx = π · 8 = 8π ≈ 25.132741Answer

✓ The volume explorer above shows the same total.

Example 3: washers

Spin the region between y = x and y = x² about the x-axis.

R = x, r = x² (x ≥ x² on 0 ≤ x ≤ 1)Outer and inner radiiπ ∫₀¹ (x² − x⁴) dx = π(1/3 − 1/5)Square each radius2π/15 ≈ 0.418879Answer

✓ Using (R − r)² instead would give π/30, a quarter of the correct answer.

Example 4: shells

Spin the region under y = 2x − x², 0 ≤ x ≤ 2, about the y-axis.

Shell radius x, height 2x − x²Vertical strips2π ∫₀² (2x² − x³) dx = 2π(16/3 − 4)Integrate8π/3 ≈ 8.37758Answer

✓ Washers would need x in terms of y, x = 1 ± √(1 − y): possible, but messier.

Example 5: a spring

A force of 10 N stretches a spring 0.2 m. How much work stretches it 0.3 m from its natural length?

10 = k · 0.2, so k = 50 N/mHooke’s law∫₀^0.3 50x dx = 25 · 0.09 = 2.25 JAnswer

✓ ½kd² = ½ · 50 · 0.3² = 2.25 J.

Example 6: pumping a tank

A cylindrical tank of radius 1 m and height 2 m is full of water. How much work pumps all the water out over the top?

A layer at height y has volume π · 1² · dySlice horizontallyIts weight is 9800π dy N, lifted 2 − y mForce and distance∫₀² 9800π(2 − y) dy = 9800π · 2Integrate19,600π ≈ 61,575 JAnswer

✓ Same as lifting the whole 19,600π N of water by 1 m, the distance its centre must rise.

Section VIIICommon mistakes

Bottom minus top. A negative area means the curves were swapped.
Ignoring a crossing. Split the interval where the curves cross.
(R − r)² for washers. It is R² − r².
Forgetting π, or 2π for shells.
Mixing units in work problems. Use metres and newtons to get joules.

Section IXPractice problems

Try each one on paper first, then open the answer to check your steps.

Find the area between y = 2x and y = x². Area
Show answer and steps
2x = x² gives x = 0 and x = 2Intersections2x ≥ x² on 0 ≤ x ≤ 2Which is on top∫₀² (2x − x²) dx = 4 − 8/3 = 4/3Answer
✓ A midpoint sum with n = 1000 gives 1.333334.
Find the area between y = x and y = x³ for 0 ≤ x ≤ 1. Area
Show answer and steps
x ≥ x³ on 0 ≤ x ≤ 1Which is on top∫₀¹ (x − x³) dx = 1/2 − 1/4 = 1/4Answer
✓ On −1 ≤ x ≤ 1 the curves swap places at 0, so the total area there is 2 · 1/4 = 1/2, not the signed integral 0.
Find the area between y = cos x and y = sin x for 0 ≤ x ≤ π/4. Area
Show answer and steps
cos x ≥ sin x on this intervalThey meet at π/4[sin x + cos x] from 0 to π/4 = √2 − 1Answer
✓ √2 − 1 ≈ 0.414214.
Spin the region under y = x, 0 ≤ x ≤ 3, about the x-axis. Find the volume. Disks
Show answer and steps
R = xDisk radiusπ ∫₀³ x² dx = π · 9 = 9πAnswer
✓ The solid is a cone of radius 3 and height 3, and ⅓πr²h = ⅓π · 9 · 3 = 9π.
Spin the region under y = x², 0 ≤ x ≤ 1, about the x-axis. Find the volume. Disks
Show answer and steps
R = x²Disk radiusπ ∫₀¹ x⁴ dx = π/5Answer
✓ π/5 ≈ 0.628319.
Spin the region between y = √x and y = x (0 ≤ x ≤ 1) about the x-axis. Washers
Show answer and steps
R = √x, r = x√x ≥ x on this intervalπ ∫₀¹ (x − x²) dx = π(1/2 − 1/3) = π/6Answer
✓ π/6 ≈ 0.523599. Square each radius first: π(R − r)² would be wrong.
Spin the region under y = x², 0 ≤ x ≤ 1, about the y-axis. Use shells. Shells
Show answer and steps
Shell radius x, height x²Vertical strips, spun about the y-axis2π ∫₀¹ x³ dx = 2π/4 = π/2Answer
✓ Washers in y give the same answer: π∫₀¹ (1 − y) dy = π/2.
A spring has constant k = 40 N/m. How much work stretches it from 0.1 m to 0.3 m past its natural length? Work
Show answer and steps
F(x) = 40xHooke’s law∫_0.1^0.3 40x dx = 20(0.09 − 0.01) = 1.6 JAnswer
✓ The first 0.1 m costs only 0.2 J; the spring gets harder to stretch as it goes.
Find the average value of f(x) = 1/x on 1 ≤ x ≤ e. Average value
Show answer and steps
1 over e − 1 ∫₁^e 1 over x dx = 1 over e − 1Integral ÷ width1/(e − 1) ≈ 0.581977Answer
✓ The answer lies between the smallest value 1/e ≈ 0.368 and the largest value 1.

Section XSelf-guided quiz: eight quick questions

Choose one answer for each question, then press “Check my answers.” Your best score is saved only in this browser.

1. If f(x) ≥ g(x) on a ≤ x ≤ b, the area between the curves is:

∫ (f + g) dx∫ f(x) g(x) dx∫ (f − g) dxf(b) − g(a)

2. What is the area between y = x and y = x²?

1/61/21/35/6

3. Spinning y = f(x), a ≤ x ≤ b, about the x-axis gives a volume of:

∫ f(x) dx2π ∫ x f(x) dxπ ∫ f′(x) dxπ ∫ f(x)² dx

4. A washer has outer radius R and inner radius r. Its area is:

π(R − r)²π(R² − r²)2π(R − r)πR²r²

5. Spinning y = x, 0 ≤ x ≤ 3, about the x-axis makes a cone. Its volume is:

3π9π27π9π/2

6. By shells, spinning the region under y = f(x), a ≤ x ≤ b (a ≥ 0), about the y-axis gives:

π ∫ x² f(x) dx∫ f(x)² dx2π ∫ f(x) dx2π ∫ x f(x) dx

7. A spring with k = 100 N/m is stretched 0.2 m from its natural length. The work done is:

2 J20 J4 J1 J

8. Archimedes proved that a sphere’s volume, compared with the cylinder around it, is:

1/23/42/31/3

Section XIArchimedes, Kepler and Cavalieri

Archimedes, in the first book of On the Sphere and Cylinder, showed that a sphere’s volume is two-thirds that of the cylinder around it, and its surface is two-thirds of the cylinder’s surface, bases included. He thought this his finest result and asked for a sphere and cylinder to be carved on his tomb. In 75 BC Cicero, in Sicily, went looking for the tomb and found it overgrown with brambles, recognising it by a small column carrying the figure of a sphere and a cylinder.

Johannes Kepler noticed at his second wedding that wine merchants measured barrels with a rod pushed in diagonally through the bung-hole, and wondered how that could work. The result was Nova stereometria doliorum (New stereometry of wine barrels, Linz, 1615), a study of volumes of solids of revolution that, building on Archimedes, cut solids into “indivisibles”.

Bonaventura Cavalieri (1598–1647) developed the method of indivisibles in his Geometria indivisibilibus of 1635. Cavalieri’s principle says that if two solids lie between the same pair of parallel planes and every plane parallel to them cuts both in sections of equal area, then the solids have equal volumes. That is the slicing idea behind V = ∫ A(x) dx. His method was not rigorous and was widely attacked, notably by Paul Guldin, but calculus later put it on firm ground.

NotesSources and credits

J. J. O’Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews: Archimedes of Syracuse (biography), on On the Sphere and Cylinder, the two-thirds ratios, the tomb and Cicero’s account.
MacTutor: Johannes Kepler (biography), on the wine barrels and Nova stereometria doliorum (1615).
MacTutor: Bonaventura Cavalieri (biography), on the 1635 Geometria indivisibilibus, Cavalieri’s principles and Guldin’s criticism.

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