What are area between two curves?

If f(x) ≥ g(x) on [a, b], the area between them is

NESA 2021 HSC Q15-style practice: Find the area of the region bounded by the curves y = x^2 and y = 2 x.

Find the points of intersection: x^2 = 2 x gives x(x - 2) = 0, so x = 0 and x = 2. On the interval [0, 2], the line y = 2 x lies above the parabola y = x^2. Area = _0^2 (2 x - x^2) \, dx = [ x^2 - (x^3)/(3) ]_0^2 = 4 - (8)/(3) = (4)/(3) square units. Markers reward finding the intersections, identifying the upper curve, and evaluating the definite integral correctly.

§-Syllabus dot point

NSWMaths AdvancedSyllabus dot point

How do we use definite integrals to compute areas between curves and volumes of solids of revolution?

Calculate the area under a curve, the area between two curves, and the volume of a solid of revolution about the $x$ or $y$ axis

A focused answer to the HSC Maths Advanced dot point on areas and volumes via integration. Areas under and between curves, the disk method for volumes of revolution about the $x$ and $y$ axes.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to interpret a definite integral as an area, extend that to the area between two curves, and apply the disk method to compute the volume of a solid formed by rotating a region about the $x$ or $y$ axis.

Note

Think of painting a fence panel. To find how much paint you need, you work out the area of the shape, and if the panel is the gap between two curved edges you subtract the lower edge from the upper edge. That is exactly what an area integral does: it adds up the little vertical strips of the shape. A volume of revolution is like spinning that flat shape on a potter's wheel: the shape whirls around a straight line and sweeps out a solid, the way spinning a flat cardboard rectangle traces out a cylinder or spinning a triangle traces out a cone. To find how much that solid holds, you imagine slicing it into a stack of thin coins, work out the area of each circular face, and add the coins up. That adding-up of coins is the integral, and every coin is a circle, which is why the radius gets squared and a factor of $\pi$ rides along.

The answer

If $f(x) \geq 0$ on $[a, b]$ , then the area between $y = f(x)$ and the $x$ -axis is

A = \int_a^b f(x) \, dx.

If $f(x) < 0$ on part of the interval, the integral subtracts that part. To get the geometric area (always positive), split the integral at zeros of $f$ and take absolute values:

A = \int_a^b |f(x)| \, dx.

Area between two curves

If $f(x) \geq g(x)$ on $[a, b]$ , the area between them is

A = \int_a^b (f(x) - g(x)) \, dx.

Find the intersection points by solving $f(x) = g(x)$ and use them as your limits. If the curves cross inside the interval, split the integral at each crossing and use the correct ordering of "upper minus lower" on each piece.

The reason "upper minus lower" works regardless of the axis is worth understanding, because it removes the need to worry about whether the curves are above or below the $x$ -axis. The vertical gap between the two curves at a given $x$ is $f(x) - g(x)$ whether both are positive, both negative, or straddling the axis: any constant you might add to shift both curves up cancels in the difference. So you never split at the $x$ -axis for an area-between-curves problem, only at the points where the two curves themselves cross.

Area between a curve and the $y$ -axis

If the boundary is described as $x = h(y)$ , integrate with respect to $y$ :

A = \int_c^d h(y) \, dy.

Volume of revolution about the $x$ -axis

Rotating the region under $y = f(x)$ between $x = a$ and $x = b$ about the $x$ -axis produces a solid whose cross sections perpendicular to the axis are disks of radius $f(x)$ . The volume is

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

The $[f(x)]^2$ is the part students drop. It is there because the cross section is a circle of radius $f(x)$ , and a circle's area is $\pi r^2$ , so the radius gets squared before integrating, not after. This is also why you cannot pull the $\pi$ inside or skip it: every cross section is a full disk, and $\pi$ is the area of a unit-radius one.

The volume of revolution, stage by stage

The disk method is best seen as a film: a flat region spins about an axis and sweeps out a solid, which you then slice back into disks to add up. The four stages below build exactly that picture for rotation about the $x$ -axis.

Stage 1, draw the region. Shade the region bounded by $y = f(x)$ , the $x$ -axis and the lines $x = a$ and $x = b$ . This flat region is what gets rotated; everything else follows from it.

Stage 2, take a representative disk. Picture a thin vertical strip of the region at a typical $x$ . When the region spins, that strip sweeps out a thin disk whose radius is the height of the curve, $r = f(x)$ , and whose thickness is $dx$ . Its circular face is the ellipse straddling the axis.

Stage 3, sweep the whole region. A full turn about the $x$ -axis sends every strip around, sweeping the region into a solid. Its outline is the curve above the axis and its mirror image below; the far end is a circular cap of radius $f(b)$ .

Stage 4, sum the disks. Fill the solid with these thin disks. One disk has volume $\pi r^2 \, dx = \pi [f(x)]^2 \, dx$ (area of a circle times thickness), and adding them all from $a$ to $b$ turns the sum into an integral:

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

Volume of revolution about the $y$ -axis

Rotating the region between $x = h(y)$ and the $y$ -axis, between $y = c$ and $y = d$ , about the $y$ -axis gives

V = \pi \int_c^d [h(y)]^2 \, dy.

You must invert the relation to express $x$ as a function of $y$ before integrating.

Volume between two curves

If a region is bounded by $y = f(x)$ above and $y = g(x)$ below, rotating about the $x$ -axis gives a washer with outer radius $f(x)$ and inner radius $g(x)$ :

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

Note that this is not $\pi \int (f - g)^2$ . The reason is geometric: each cross section is an annulus (a washer), whose area is the big circle minus the small hole, $\pi R^2 - \pi r^2$ . That is $\pi(R^2 - r^2)$ , the difference of the squared radii. Squaring the gap $(f - g)$ would measure the area of a disk whose radius is the gap, which is a completely different solid.

How exam questions ask about areas and volumes

The wording pins down both the method and the variable you integrate in:

"Find the area bounded by / enclosed by the curve and the $x$ -axis." A single curve against the axis: integrate $f(x)$ , but if the curve dips below the axis, split at its roots and take magnitudes so the area is positive.
"Find the area between / enclosed by the two curves." Solve $f(x) = g(x)$ for the limits, then integrate upper minus lower. Sketch first so you know which curve is on top.
"The region is rotated about the $x$ -axis." Disk method in $x$ : $V = \pi \int [f(x)]^2\,dx$ , with $x$ -limits.
"The region is rotated about the $y$ -axis." Disk method in $y$ : rearrange to $x = h(y)$ first, then $V = \pi \int [h(y)]^2\,dy$ , with $y$ -limits. The single most common error here is forgetting to invert and convert the limits to $y$ -values.
"Find the volume of the solid between $y = f(x)$ and $y = g(x)$ when rotated about the $x$ -axis." Washer method: $V = \pi \int ([f]^2 - [g]^2)\,dx$ , outer radius squared minus inner radius squared.
"Give your answer in exact form" or no rounding instruction. Leave $\pi$ , surds and fractions in place; do not reach for the calculator.

The axis of rotation decides the variable of integration every time: rotate about the $x$ -axis and you integrate in $x$ ; rotate about the $y$ -axis and you integrate in $y$ . Get that pairing wrong and the whole integral is set up incorrectly.

Worked example

This works the three standard shapes in turn: an area between two curves, a volume about the $y$ -axis (which needs inverting), and a washer between two curves.

Find the area between $y = x^3$ and $y = x$

Find the limits and decide which curve is upper: Find the area between $y = x^3$ and $y = x$ for $x \in [0, 1]$ . First decide which curve is upper. Testing $x = 0.5$ gives $x = 0.5$ and $x^3 = 0.125$ , so $y = x$ is above $y = x^3$ on this interval. The limits $0$ and $1$ are given (they are also where the curves meet, since $x = x^3$ at $x = 0, 1$ ).

Integrate upper minus lower:

A = \int_0^1 (x - x^3) \, dx = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \text{ square units}.

Find the volume about the $y$ -axis (invert first)

The region bounded by $y = x^2$ , $y = 4$ and the $y$ -axis (first quadrant) is rotated about the $y$ -axis. Because the rotation is about the $y$ -axis, integrate in $y$ , so express the radius as a function of $y$ : from $y = x^2$ , the radius is $x = \sqrt{y}$ , so $x^2 = y$ . The $y$ -limits run from $0$ (the origin) to $4$ (the line $y = 4$ ).

V = \pi \int_0^4 [\,h(y)\,]^2 \, dy = \pi \int_0^4 y \, dy = \pi \left[ \frac{y^2}{2} \right]_0^4 = 8 \pi \text{ cubic units}.

Find the washer volume between two curves

The region between $y = \sqrt{x}$ (outer) and $y = x$ (inner) from $x = 0$ to $x = 1$ is rotated about the $x$ -axis. Here $\sqrt{x} \geq x$ on $[0, 1]$ , so the outer radius is $\sqrt{x}$ and the inner radius is $x$ , and the cross section is a washer. Use the difference of squares, never the square of the difference:

V = \pi \int_0^1 \left( (\sqrt{x})^2 - x^2 \right) dx = \pi \int_0^1 (x - x^2)\,dx = \pi \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \pi \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{\pi}{6} \text{ cubic units}.

As a check that the washer is needed, the solid from the outer curve alone would be $\pi \int_0^1 x\,dx = \frac{\pi}{2}$ , and the solid from the inner curve alone $\pi \int_0^1 x^2\,dx = \frac{\pi}{3}$ ; the washer is their difference $\frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$ , exactly as the formula gives.

Common traps

Forgetting the absolute value: A definite integral can be negative if the curve dips below the axis. Area is non-negative, so split or take absolute values.
Squaring the difference instead of differencing the squares: Volume of a washer uses $[f(x)]^2 - [g(x)]^2$ , not $(f(x) - g(x))^2$ .
Integrating with respect to the wrong variable: Rotation about the $x$ -axis pairs with $\int f(x)^2 \, dx$ . Rotation about the $y$ -axis pairs with $\int h(y)^2 \, dy$ .
Missing intersections: If the curves cross inside your interval, the upper and lower roles swap. Split the integral and re-test which curve is on top in each subinterval.
Dropping $\pi$: A volume of revolution always carries a factor of $\pi$ . Forgetting it costs marks.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC Q154 marksFind the area of the region bounded by the curves

y = x^2

and

y = 2 x

Show worked answer →

Find the points of intersection: $x^2 = 2 x$ gives $x(x - 2) = 0$ , so $x = 0$ and $x = 2$ .

On the interval $[0, 2]$ , the line $y = 2 x$ lies above the parabola $y = x^2$ .

Area $= \int_0^2 (2 x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$ square units.

Markers reward finding the intersections, identifying the upper curve, and evaluating the definite integral correctly.

2019 HSC Q144 marksThe region bounded by

y = \sqrt{x}

, the

x

-axis and the line

x = 4

is rotated about the

x

-axis. Find the volume of the solid formed.

Show worked answer →

Use the disk method. Volume $= \pi \int_a^b y^2 \, dx$ .

With $y = \sqrt{x}$ , $y^2 = x$ .

$V = \pi \int_0^4 x \, dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \cdot 8 = 8 \pi$ cubic units.

Markers expect the disk-method formula, the correct $y^2$ , evaluation between $0$ and $4$ , and the answer left in exact form with $\pi$ .

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksFind the area between the curve

y = 3x^2

and the

x

-axis from

x = 1

x = 2

Show worked solution →

Set up the area integral. The curve is positive on $[1, 2]$ , so the area is the definite integral of $y$ :

A = \int_1^2 3x^2 \, dx.

Integrate and evaluate.

A = \left[ x^3 \right]_1^2 = 2^3 - 1^3 = 8 - 1 = 7 \text{ square units}.

Marker's note: one mark for the correct integral $\int_1^2 3x^2\,dx$ with the antiderivative $x^3$ , one for evaluating to $7$ square units. Writing the answer without "square units" is a common minor loss.

foundation3 marksFind the area of the region enclosed between the line

y = x

and the parabola

y = x^2

Show worked solution →

Find the intersection points for the limits. Solve $x = x^2$ :

x^2 - x = 0 \quad\Rightarrow\quad x(x - 1) = 0 \quad\Rightarrow\quad x = 0 \text{ or } x = 1.

Decide which curve is upper. Testing $x = \tfrac{1}{2}$ gives $y = \tfrac{1}{2}$ for the line and $y = \tfrac{1}{4}$ for the parabola, so $y = x$ is above $y = x^2$ on $[0, 1]$ .

Integrate upper minus lower.

A = \int_0^1 (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \text{ square units}.

Marker's note: one mark for the limits $x = 0$ and $x = 1$ , one for the correct "upper minus lower" integrand $x - x^2$ , one for evaluating to $\frac{1}{6}$ . Integrating $x^2 - x$ (wrong order) gives $-\frac{1}{6}$ and loses the final mark.

foundation2 marksThe region under the line

y = 2x

from

x = 0

x = 3

is rotated about the

x

-axis. Find the volume of the cone formed.

Show worked solution →

Apply the disk method. Rotating about the $x$ -axis gives $V = \pi \int_a^b [f(x)]^2 \, dx$ with $f(x) = 2x$ , so $[f(x)]^2 = 4x^2$ :

V = \pi \int_0^3 4x^2 \, dx.

Integrate and evaluate.

V = \pi \left[ \frac{4x^3}{3} \right]_0^3 = \pi \cdot \frac{4 \times 27}{3} = 36\pi \text{ cubic units}.

Marker's note: one mark for the set-up $\pi \int_0^3 (2x)^2\,dx$ with the radius squared, one for evaluating to $36\pi$ cubic units. Forgetting to square the $2$ (using $2x^2$ instead of $4x^2$ ) halves the answer and loses the mark.

core4 marksThe parabola

y = x^2 - 4

is graphed. It has its vertex at

(0, -4)

, cuts the

x

-axis at

x = -2

and

x = 2

, and lies entirely below the

x

-axis for

-2 < x < 2

. Find the area of the region enclosed between the curve and the

x

-axis.

Show worked solution →

Recognise the region sits below the axis. Between $x = -2$ and $x = 2$ the curve is below the $x$ -axis, so the plain definite integral would come out negative. Area is non-negative, so take the magnitude:

A = \left| \int_{-2}^{2} (x^2 - 4) \, dx \right| = \int_{-2}^{2} (4 - x^2) \, dx.

Integrate the flipped integrand.

A = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2}.

Evaluate at each limit.

\left( 4(2) - \frac{8}{3} \right) - \left( 4(-2) - \frac{-8}{3} \right) = \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \text{ square units}.

Marker's note: one mark for recognising the region is below the axis and using $|\cdot|$ (or integrating $4 - x^2$ ), one for the correct antiderivative, one for correct substitution of both limits, one for $\frac{32}{3}$ square units. Reporting $-\frac{32}{3}$ or forgetting the sign issue is the trap the graph is drawn to expose.

core4 marksThe region in the first quadrant bounded by the curve

y = x^2

, the line

y = 9

and the

y

-axis is rotated about the

y

-axis. Find the exact volume of the solid formed.

Show worked solution →

Rotation about the $y$ -axis means integrate in $y$ . Use $V = \pi \int_c^d [h(y)]^2 \, dy$ , so first express the radius $x$ as a function of $y$ . From $y = x^2$ (first quadrant),

x = \sqrt{y} \quad\Rightarrow\quad x^2 = y = [h(y)]^2.

Set the $y$ -limits. The region runs from $y = 0$ (the origin) up to $y = 9$ (the line).

Integrate.

V = \pi \int_0^9 y \, dy = \pi \left[ \frac{y^2}{2} \right]_0^9 = \pi \cdot \frac{81}{2} = \frac{81\pi}{2} \text{ cubic units}.

Marker's note: one mark for choosing to integrate in $y$ , one for inverting to $[h(y)]^2 = y$ , one for the correct limits $0$ to $9$ , one for $\frac{81\pi}{2}$ cubic units. The classic error is leaving the integral in $x$ and rotating as if about the $x$ -axis.

exam5 marksA region

R

is bounded by the curve

y = \sqrt{x}

and the line

y = \dfrac{x}{2}

. (a) Show that the curve and the line meet at

x = 0

and

x = 4

. (b) Show that the area of

R

\dfrac{4}{3}

square units. (c) Hence, or otherwise, find the exact volume of the solid formed when

R

is rotated about the

x

-axis.

Show worked solution →

Part (a): solve $\sqrt{x} = \dfrac{x}{2}$ for the intersections. Square both sides:

x = \frac{x^2}{4} \quad\Rightarrow\quad 4x = x^2 \quad\Rightarrow\quad x^2 - 4x = 0 \quad\Rightarrow\quad x(x - 4) = 0,

so $x = 0$ and $x = 4$ , as required.

Part (b): integrate upper minus lower. Testing $x = 1$ gives $\sqrt{x} = 1$ and $\dfrac{x}{2} = \dfrac{1}{2}$ , so $y = \sqrt{x}$ is the upper curve on $[0, 4]$ :

A = \int_0^4 \left( \sqrt{x} - \frac{x}{2} \right) dx = \left[ \frac{2}{3} x^{3/2} - \frac{x^2}{4} \right]_0^4.

Substitute $x = 4$ (using $4^{3/2} = 8$ ):

A = \frac{2}{3}(8) - \frac{16}{4} = \frac{16}{3} - 4 = \frac{16}{3} - \frac{12}{3} = \frac{4}{3} \text{ square units}.

Part (c): rotate $R$ about the $x$ -axis using the washer method. The outer radius is the upper curve $\sqrt{x}$ and the inner radius is the line $\dfrac{x}{2}$ , so use the difference of squared radii:

V = \pi \int_0^4 \left( (\sqrt{x})^2 - \left( \frac{x}{2} \right)^2 \right) dx = \pi \int_0^4 \left( x - \frac{x^2}{4} \right) dx.

Integrate and evaluate:

V = \pi \left[ \frac{x^2}{2} - \frac{x^3}{12} \right]_0^4 = \pi \left( \frac{16}{2} - \frac{64}{12} \right) = \pi \left( 8 - \frac{16}{3} \right) = \pi \cdot \frac{8}{3} = \frac{8\pi}{3} \text{ cubic units}.

Marker's note: one mark for part (a) solving to $x = 0, 4$ ; one mark for the correct "upper minus lower" integrand in (b) and one for evaluating to $\frac{4}{3}$ ; one mark in (c) for the washer set-up with squared radii (not the square of the difference) and one for $\frac{8\pi}{3}$ cubic units. Writing $\pi\int(\sqrt{x} - \tfrac{x}{2})^2\,dx$ is the standard washer trap and scores neither (c) mark.

exam5 marksA designer models a bowl by rotating the region under the curve

y = \dfrac{1}{2}x^2

, between

x = 0

and

x = 4

, about the

y

-axis. All lengths are in centimetres. (a) Show that, written in terms of

y

, the squared radius of the bowl satisfies

x^2 = 2y

. (b) Find the depth of the bowl, that is the value of

y

where

x = 4

. (c) Find the exact capacity of the bowl in cubic centimetres.

Show worked solution →

Part (a): invert the curve. Rotation is about the $y$ -axis, so the radius must be a function of $y$ . From $y = \dfrac{1}{2}x^2$ ,

2y = x^2 \quad\Rightarrow\quad x^2 = 2y,

as required.

Part (b): find the depth. The rim of the bowl is at $x = 4$ , so

y = \frac{1}{2}(4)^2 = \frac{1}{2}(16) = 8 \text{ cm}.

The bowl is $8$ cm deep.

Part (c): the capacity is the volume of revolution in $y$ . Using $V = \pi \int_c^d x^2 \, dy$ with $x^2 = 2y$ and $y$ running from $0$ to $8$ :

V = \pi \int_0^8 2y \, dy = \pi \left[ y^2 \right]_0^8 = \pi (64 - 0) = 64\pi \text{ cm}^3.

Marker's note: one mark for inverting to $x^2 = 2y$ in (a); one mark for the depth $y = 8$ cm in (b); one mark for the volume set-up $\pi\int_0^8 2y\,dy$ with $y$ -limits, one for the antiderivative $y^2$ , and one for $64\pi$ cm $^3$ . Integrating in $x$ , or using $x$ -limits $0$ to $4$ with the $y$ -integrand, are the two common set-up errors.

What this dot point is asking

The answer

Area between two curves

Area between a curve and the yyy-axis

Volume of revolution about the xxx-axis

The volume of revolution, stage by stage

Volume of revolution about the yyy-axis

Volume between two curves

How exam questions ask about areas and volumes

Exam-style practice questions

Practice questions

Area between a curve and the $y$ -axis

Volume of revolution about the $x$ -axis

Volume of revolution about the $y$ -axis