# Area between curves, volumes of revolution

> IB Mathematics: Applications and Interpretation HL · Calculus
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u4-area-between-curves-volumes-of/

This module covers core applications of definite integration for IB AI HL: calculating area of regions bounded by one or more curves, and finding volumes of solids formed by rotating regions around the x-axis.

**Prerequisites:** [Definite integration](https://www.owlsprep.com/study/ib-math-ai-hl-u4-definite-integration/); Finding intersection points of functions

## Learning objectives

- Calculate the area between a curve and a coordinate axis
- Find the total area between two intersecting curves
- Calculate volumes of revolution around the x-axis using disk/washer method
- Identify and avoid common exam traps in area/volume problems

## Area between a curve and a coordinate axis

**Area between curve and axis** — 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.

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

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 = $\left|\int_0^2 (x^2 - 4) dx\right| + \left|\int_2^3 (x^2 - 4) dx\right|$
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. $$\frac{16}{3} + \frac{7}{3} = \frac{23}{3} \approx 7.67 \text{ square units}$$

## Area between two curves

**Area between two curves** — 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.

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

> **tip**
>
> Always test a point between the two intersection points to confirm which function is upper. Never assume the higher function is the one written first.

**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.

## Volumes of revolution around the x-axis

**Disk and Washer Method** — 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.

*Notation:* 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

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 $\pi r^2 = \pi [f(x)]^2$. 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 $\pi$ for final volume:
10. $$V = \frac{26}{3}\pi \approx 27.2 \text{ cubic units}$$

**Check your understanding**

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:

1. Which formula is correct?

   - A: $\pi \int_a^b (f(x) - g(x))^2 dx$
   - B: $\pi \int_a^b (f(x)^2 - g(x)^2) dx$
   - C: $\pi \int_a^b f(x)^2 dx - \pi \int_b^a g(x)^2 dx$
   - D: $\int_a^b \pi f(x)^2 g(x)^2 dx$

   *Why:* 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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to split the integral and take absolute value for negative regions when calculating area.
  - Why it fails: The definite integral gives net signed area, which can be much smaller than total area (even zero if regions cancel out).
  - Correct: Find all x-intercepts or intersection points between bounds, split the integral, add the absolute value of each sub-integral.
- **Wrong:** Reversing upper and lower functions when calculating area between curves, leading to a negative answer.
  - Why it fails: Area is always positive; exam markers will deduct marks for negative final answers even if integration is correct.
  - Correct: Confirm which function is larger with a test point, subtract lower from upper, or take the absolute value of the final result.
- **Wrong:** Forgetting to multiply by $\pi$ in the volume of revolution formula.
  - Why it fails: The formula is derived from the area of a circle ($A = \pi r^2$), so omitting $\pi$ gives an incorrect answer.
  - Correct: Always write the $\pi$ factor explicitly when setting up your volume integral.
- **Wrong:** Using the x-axis rotation formula for rotation around the y-axis.
  - Why it fails: The formula $V = \pi \int [f(x)]^2 dx$ only applies to rotation around the x-axis.
  - Correct: If rotating around the y-axis, rearrange your function to $x = f(y)$ and integrate over y instead.
- **Wrong:** Not splitting the integral when curves cross between bounds.
  - Why it fails: If the upper and lower functions swap places, the integral will give net area instead of total bounded area.
  - Correct: Find all intersection points in your interval before setting up the integral, split into sub-regions.

## Cheatsheet

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

## 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.

- [First order differential equations: Euler's method](https://www.owlsprep.com/study/ib-math-ai-hl-u4-first-order-differential-equations-euler/)
- [Separation of variables for differential equations](https://www.owlsprep.com/study/ib-math-ai-hl-u4-separation-of-variables-for-differential/)
- [Differential equations for growth, decay and logistic models](https://www.owlsprep.com/study/ib-math-ai-hl-u4-differential-equations-for-growth-decay/)

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