Area between curves, volumes of revolution

IB Mathematics: Applications and Interpretation HL· 40 min read

1. Area between a curve and a coordinate axis★★☆☆☆⏱ 15 min

📘 Definition

Area between curve and axis

\text{Area} = \int_a^b |f(x)| dx

For a continuous function $y = f (x)$ , the total area between the curve, the x-axis, $x = a$ and $x = b$ is the integral of the absolute value of $f (x)$ over the interval. If $f (x)$ crosses the x-axis, split the integral at intercepts.

The definite integral gives the net signed area, where regions below the x-axis count as negative. Total area (what exam questions almost always ask for) requires adding the magnitude of each positive and negative region.

📐 Worked Example

Find the total area bounded by $f (x) = x^{2} - 4$ , the x-axis, $x = 0$ and $x = 3$ .

1
Find x-intercepts in the interval:
2
$x^2 - 4 = 0 \implies x = 2 \quad (x=-2 \text{ is outside } [0,3])$
3
$f (x)$ is negative on $[0, 2]$ and positive on $[2, 3]$ , so split the integral and add absolute values:
4
Area = $\int_{0}^{2} (x^{2} - 4) d x + \int_{2}^{3} (x^{2} - 4) d x$
5
Evaluate the first integral:
6
$\int_0^2 (x^2 - 4) dx = \left[\frac{x^3}{3} - 4x\right]_0^2 = \frac{8}{3} - 8 = -\frac{16}{3}$
7
Absolute value: $\frac{16}{3}$ . Evaluate the second integral:
8
$\int_2^3 (x^2 - 4) dx = \left[\frac{x^3}{3} - 4x\right]_2^3 = (-3) - \left(-\frac{16}{3}\right) = \frac{7}{3}$
9
Add areas for total area:
10
$\frac{16}{3} + \frac{7}{3} = \frac{23}{3} \approx 7.67 \text{ square units}$

2. Area between two curves★★★☆☆⏱ 20 min

📘 Definition

Area between two curves

\text{Area} = \int_a^b \left(f_{\text{upper}}(x) - f_{\text{lower}}(x)\right) dx

For two continuous functions, the total area between them from $x = a$ to $x = b$ is the integral of the upper function minus the lower function over the interval. If curves intersect, split at intersection points.

📐 Worked Example

Find the total area bounded by $y = x^{2}$ and $y = x + 2$ .

1
Find intersection points by setting functions equal:
2
$x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0$
3
Intersections at $x = - 1$ and $x = 2$ , so our interval is $[- 1, 2]$ .
4
Test the point $x = 0$ : $y = x + 2 = 2$ , $y = x^{2} = 0$ , so $x + 2$ is the upper function.
5
Set up the area integral:
6
$\text{Area} = \int_{-1}^2 \left((x+2) - x^2\right) dx$
7
Integrate term-by-term:
8
$\int (x + 2 - x^2) dx = \frac{x^2}{2} + 2x - \frac{x^3}{3}$
9
Evaluate at bounds:
10
$\left(\frac{4}{2} + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{27}{6} = 4.5$
11
Total area is $4.5$ square units.

3. Volumes of revolution around the x-axis★★★☆☆⏱ 25 min

✓ Calculator OK

📘 Definition

Disk and Washer Method

Disk: V = \pi \int_a^b [f(x)]^2 dx, Washer: V = \pi \int_a^b \left([f(x)]^2 - [g(x)]^2\right) dx

For rotation of a region around the x-axis, volume is found by summing infinitely thin circular disks. The disk method is for regions between a curve and the x-axis; the washer method subtracts the inner hole for regions between two curves.

The formula comes from the area of a circle: each disk at position $x$ has radius equal to $f (x)$ , so cross-sectional area is $π r^{2} = π [f (x)]^{2}$ . Integrating along x gives total volume.

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = x + 1$ , the x-axis, $x = 0$ and $x = 2$ is rotated 360° around the x-axis.

1
We use the disk method for a region between curve and x-axis:
2
$V = \pi \int_0^2 [f(x)]^2 dx = \pi \int_0^2 (x+1)^2 dx$
3
Expand the integrand:
4
$(x+1)^2 = x^2 + 2x + 1$
5
Integrate term-by-term:
6
$\int_0^2 (x^2 + 2x +1) dx = \left[\frac{x^3}{3} + x^2 + x\right]_0^2$
7
Evaluate at bounds:
8
$\left(\frac{8}{3} + 4 + 2\right) - 0 = \frac{26}{3}$
9
Multiply by $π$ for final volume:
10
$V = \frac{26}{3}\pi \approx 27.2 \text{ cubic units}$

✓ Quick check

Select the correct formula for the volume when the region between $y = f (x)$ and $y = g (x)$ (where $f (x) > g (x) > 0$ for $a \leq x \leq b$ ) is rotated 360° around the x-axis:

Which formula is correct?
- A: $π \int_{a}^{b} (f (x) - g (x))^{2} d x$
- B: $π \int_{a}^{b} (f (x)^{2} - g (x)^{2}) d x$
- C: $π \int_{a}^{b} f (x)^{2} d x - π \int_{b}^{a} g (x)^{2} d x$
- D: $\int_{a}^{b} π f (x)^{2} g (x)^{2} d x$
Reveal answer
B —
Option B is the correct washer method. It subtracts the volume of the inner hole (radius $g (x)$ ) from the outer solid (radius $f (x)$ ). Option A is wrong because it squares the difference of radii, not taking the difference of squared radii.

4. Common Pitfalls

Wrong move:

Forgetting to split the integral and take absolute value for negative regions when calculating area.

Why:

The definite integral gives net signed area, which can be much smaller than total area (even zero if regions cancel out).

Correct move:

Find all x-intercepts or intersection points between bounds, split the integral, add the absolute value of each sub-integral.

Wrong move:

Reversing upper and lower functions when calculating area between curves, leading to a negative answer.

Why:

Area is always positive; exam markers will deduct marks for negative final answers even if integration is correct.

Correct move:

Confirm which function is larger with a test point, subtract lower from upper, or take the absolute value of the final result.

Wrong move:

Forgetting to multiply by $π$ in the volume of revolution formula.

Why:

The formula is derived from the area of a circle ( $A = π r^{2}$ ), so omitting $π$ gives an incorrect answer.

Correct move:

Always write the $π$ factor explicitly when setting up your volume integral.

Wrong move:

Using the x-axis rotation formula for rotation around the y-axis.

Why:

The formula $V = π \int [f (x)]^{2} d x$ only applies to rotation around the x-axis.

Correct move:

If rotating around the y-axis, rearrange your function to $x = f (y)$ and integrate over y instead.

Wrong move:

Not splitting the integral when curves cross between bounds.

Why:

If the upper and lower functions swap places, the integral will give net area instead of total bounded area.

Correct move:

Find all intersection points in your interval before setting up the integral, split into sub-regions.

5. Quick Reference Cheatsheet

Concept	Formula	Notes
Area between $y = f (x)$ and x-axis ( $a \leq x \leq b$ )	$\int_{a}^{b} ∣ f (x) ∣ d x$	Split at x-intercepts, add absolute values
Area between $y = f (x), y = g (x)$ ( $a \leq x \leq b$ )	$\int_{a}^{b} ∣ f (x) - g (x) ∣ d x$	Split at intersections, subtract lower from upper
Volume (disk, rotate around x-axis)	$V = π \int_{a}^{b} [f (x)]^{2} d x$	Region between curve and x-axis
Volume (washer, rotate around x-axis)	$V = π \int_{a}^{b} ([f]^{2} - [g]^{2}) d x$	$f$ = outer radius, $g$ = inner radius

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 2
Area between two exponential curves
2022 · 1
Volume of revolution of a cubic function
2023 · 2
Total area between intersecting curves

What's Next

Area between curves and volumes of revolution are core applied integration topics regularly tested in both Paper 1 and Paper 2 of IB AI HL. They are often combined with other topics: for example, you may be asked to find the intersection of a quadratic and an exponential function, then calculate the volume of revolution of the bounded region, or find the area between a velocity curve and the time axis for kinematics problems. Mastering these techniques builds a strong foundation for further applications of integration, including solving differential equations and modelling dynamic systems.