Applications of integration: area and volume

IB Mathematics: Analysis and Approaches HL· Syllabus 5.11: Areas, volumes of revolution· 25 min read

1. Area between a curve and the x-axis★★☆☆☆⏱ 7 min

📘 Definition

Signed Area vs Total Area

Definite integration calculates signed area: area above the x-axis is positive, area below is negative. Total geometric area is the sum of absolute values of signed areas over each interval where the function does not change sign.

When a function crosses the x-axis within your integration interval, you must split the integral into separate parts for regions above and below the axis. Failing to do this causes positive and negative areas to cancel, giving an incorrect result.

📐 Worked Example

Find the total geometric area bounded by $f (x) = x$ and the x-axis between $x = - 2$ and $x = 3$ .

1
Find where $f (x)$ crosses the x-axis: $x = 0$ , so split into intervals $[- 2, 0]$ and $[0, 3]$ . Calculate signed area on the first interval:
2
$\int_{-2}^0 x dx = \left[\frac{1}{2}x^2\right]_{-2}^0 = 0 - 2 = -2$
3
Take the absolute value of the negative signed area, add the positive area from the second interval:
4
$\text{Total Area} = \left| \int_{-2}^0 x dx \right| + \int_0^3 x dx = |-2| + \left[\frac{1}{2}x^2\right]_0^3 = 2 + 4.5 = 6.5$
5
Final total area is $\frac{13}{2}$ square units.

Exam tip:

Always check for roots of $f (x) = 0$ in your interval before integrating to find area.

2. Area between two curves★★★☆☆⏱ 8 min

📘 Definition

Area between two curves

For continuous functions $f (x) \geq g (x)$ over $[a, b]$ , the total area between the curves is $\int_{a}^{b} (f (x) - g (x)) d x$ . Bounds $a, b$ are almost always the x-coordinates of intersection of the two curves.

If the curves cross each other within the interval, split the integral at the intersection point, just like when crossing the x-axis. Always subtract the lower function from the upper function to keep the integrand positive.

📐 Worked Example

Find the area bounded by $f (x) = x^{2}$ and $g (x) = x + 2$ .

1
Find intersection points by setting the functions equal:
2
$x^2 = x+2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0$
3
Bounds are $a = - 1$ , $b = 2$ . Test to confirm $g (x) > f (x)$ over this interval, so set up the integral:
4
$\text{Area} = \int_{-1}^2 \left((x+2) - x^2\right) dx = \left[\frac{1}{2}x^2 + 2x - \frac{1}{3}x^3\right]_{-1}^2$
5
Evaluate: upper bound at $x = 2$ is $\frac{10}{3}$ , lower bound at $x = - 1$ is $- \frac{7}{6}$ :
6
$\frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{27}{6} = \frac{9}{2} = 4.5$
7
Total area is $4.5$ square units.

Exam tip:

If bounds are not given, always solve for intersection points first to earn full method marks.

3. Volume of revolution around the x-axis★★★☆☆⏱ 9 min

Rotating a region bounded by $y = f (x)$ , the x-axis, $x = a$ , and $x = b$ 360° around the x-axis creates a solid. The cross-section perpendicular to the x-axis is a circle with radius $∣ f (x) ∣$ , leading to the disk method formula:

V = \pi \int_a^b [f(x)]^2 dx

For a region between two curves rotated around the x-axis, use the washer method for hollow solids, where cross-sectional area is the area of the outer disk minus the inner disk:

V = \pi \int_a^b \left(R^2 - r^2\right) dx = \pi \int_a^b \left([f_{\text{outer}}(x)]^2 - [f_{\text{inner}}(x)]^2\right) dx

📐 Worked Example

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

1
Confirm $y = x + 1 \geq 0$ over $[0, 2]$ , so radius is $f (x) = x + 1$ . Apply the disk formula:
2
$V = \pi \int_0^2 (x+1)^2 dx = \pi \int_0^2 (x^2 + 2x + 1) dx$
3
Integrate and evaluate:
4
$\pi \left[ \frac{1}{3}x^3 + x^2 + x \right]_0^2 = \pi \left( \frac{8}{3} + 4 + 2 \right) = \frac{26}{3}\pi \approx 27.23$
5
Volume is $\frac{26}{3} π$ cubic units.

4. Volume of revolution around the y-axis★★★★☆⏱ 8 min

For rotation around the y-axis, IB expects integration with respect to y. First rewrite your function $y = f (x)$ as $x = x (y)$ , then find bounds for y. The formula is analogous to rotation around the x-axis:

V = \pi \int_c^d [x(y)]^2 dy

The shell method is not required for IB AA HL, so you will only be tested on this standard method integrating with respect to y.

📐 Worked Example

Find the volume when the region bounded by $y = x^{2}$ , $y = 0$ , and $y = 4$ is rotated 360° around the y-axis.

1
Rewrite $y = x^{2}$ as $x = y$ (positive root for $x \geq 0$ ). Bounds for y are $c = 0$ , $d = 4$ . Apply the formula:
2
$V = \pi \int_0^4 (\sqrt{y})^2 dy = \pi \int_0^4 y dy$
3
Evaluate the integral:
4
$\pi \left[ \frac{1}{2}y^2 \right]_0^4 = \pi (8 - 0) = 8\pi$
5
Volume is $8 π$ cubic units.

Exam tip:

Always double-check which axis you are rotating around, and integrate with respect to the correct variable.

5. Common Pitfalls

Wrong move:

Forgetting to split the integral when curves cross the x-axis or each other

Why:

Signed areas cancel out, resulting in a total area much smaller than the true geometric area

Correct move:

Find all crossing points in the interval, split the integral, take the absolute value of each part before adding

Wrong move:

Omitting $π$ from the volume of revolution formula

Why:

The area of a circular cross-section includes $π$ , so volume must include the constant

Correct move:

Always write $π$ outside the integral when setting up a volume calculation

Wrong move:

Squaring the difference of two radii for the washer method

Why:

$π (R - r)^{2} \neq = π (R^{2} - r^{2})$ , so this gives an incorrect volume

Correct move:

Square each radius separately, then subtract the squared values before integrating

Wrong move:

Integrating with respect to x when rotating around the y-axis

Why:

The standard IB method requires integration over y for rotation around the y-axis

Correct move:

Rearrange the function to get x in terms of y, update bounds to y-values, integrate with respect to y

Wrong move:

Subtracting the upper function from the lower function for area between curves

Why:

This gives a negative integrand, leading to negative area which is impossible for geometric area

Correct move:

Confirm which function is upper/lower over each interval, always subtract lower from upper

6. Quick Reference Cheatsheet

Calculation Type	Formula	Key Notes
Area: curve to x-axis	$\sum \int_{a_{i}}^{b_{i}} f (x) d x$	Split at sign changes of $f (x)$
Area between two curves	$\int_{a}^{b} (f_{u pp er} - f_{l o w er}) d x$	$a, b$ = intersection x-values
Volume rotation: x-axis (disk)	$V = π \int_{a}^{b} [f (x)]^{2} d x$	Cross-section perpendicular to x-axis
Volume rotation: x-axis (washer)	$V = π \int_{a}^{b} (R^{2} - r^{2}) d x$	R = outer radius, r = inner radius
Volume rotation: y-axis	$V = π \int_{c}^{d} [x (y)]^{2} d y$	Rewrite x as function of y first

7. Frequently Asked

Do I need to split integrals for area when curves cross the axis?

Yes. Integration gives signed area, so positive and negative areas will cancel if you do not split. Split at all crossing points, take absolute values, then sum.

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.

2025 · 1
Area between two quadratic curves
2024 · 2
Volume of revolution around x-axis
2023 · 1
Area crossing the x-axis

What's Next

Area and volume calculations are core integration applications that frequently appear in both Paper 1 and Paper 2 IB AA HL exams, often as part of longer multi-part problems. Mastering the sign rules and formula setup here will help you avoid losing easy method marks on exam day. This subtopic builds directly on your knowledge of definite integration, and paves the way for other integration applications like kinematics, differential equations, and modeling. Next, you can extend your skills to other common integration applications or review integration techniques for more complex functions.