---
title: Areas Between Curves and Volumes
module: Applications of Integration
moduleNumber: 5
lessonNumber: 1
order: 501
summary: >
  A definite integral computes any quantity that a limit of Riemann sums
  approximates. Applied to geometry it gives the area between two curves and the
  volume of a solid: by cross-sections, by disks and washers when the region is
  revolved, and by cylindrical shells when inverting the boundary is awkward.
topics: [Applications of Integration]
sources:
  - book: Stewart
    ref: "Ch. 5 — Applications of Integration; §5.1 Areas Between Curves"
  - book: Stewart
    ref: "§5.2 Volumes; §5.3 Volumes by Cylindrical Shells"
---

A definite integral computes any quantity that a limit of Riemann sums
approximates. Cut a geometric or physical quantity $Q$ into many thin pieces;
approximate each piece by a term $f(x_i^\ast)\,\Delta x$; sum the pieces; take
the limit. The sum becomes $\int_a^b f(x)\, \d x$, evaluated by the
[Fundamental Theorem](/calculus/integrals/the-fundamental-theorem-of-calculus).
Area under a single curve is the
[first instance](/calculus/integrals/area-and-the-definite-integral); the area
between two curves and the volume of a solid apply the same construction to a
different piece.

## Area between two curves

Consider a region $S$ bounded above by $y = f(x)$, below by $y = g(x)$, and on
the sides by $x = a$ and $x = b$, with $f(x) \ge g(x)$ throughout. Slicing $S$
into vertical strips of width $\Delta x$, the strip at $x_i^\ast$ is nearly a
rectangle of height $f(x_i^\ast) - g(x_i^\ast)$. Summing the strip areas and
passing to the limit gives a Riemann sum whose value is the integral of the gap.

> **Definition (Area between curves).** If $f$ and $g$ are continuous with
> $f(x) \ge g(x)$ on $[a, b]$, the area of the region bounded by $y = f(x)$,
> $y = g(x)$, $x = a$, and $x = b$ is
>
> $$
> A = \int_a^b \bigl[f(x) - g(x)\bigr]\, \d x = \int_a^b (y_T - y_B)\, \d x,
> $$
>
> where $y_T$ is the top boundary and $y_B$ the bottom.[^area-def]

The special case $g(x) = 0$ recovers the area under a single curve. The formula
carries no assumption that either curve is positive: the difference $y_T - y_B$
is the height of the strip whether the strip sits above the axis, below it, or
straddles it.

$$
% caption: The area is the integral of the strip height y_T - y_B; the top and
% bottom boundaries are read off a sketch, not memorized.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (-0.3,0) -- (6.4,0) node[right] {$x$};
\draw[->, black] (0,-0.3) -- (0,4.2) node[above] {$y$};
% top curve y=f(x)
\draw[black, thick] (0.4,3.6) .. controls (2.4,1.2) and (3.6,3.4) .. (5.8,3.9)
  node[right, black] {$y_T = f(x)$};
% bottom curve y=g(x)
\draw[black, thick] (0.4,1.1) .. controls (2.6,0.3) and (4.0,1.4) .. (5.8,1.3)
  node[right] {$y_B = g(x)$};
% representative strip
\filldraw[acc!12, draw=acc] (3.0,0.86) rectangle (3.28,2.55);
\draw[<->, black!70] (3.9,0.9) -- (3.9,2.5) node[midway, right] {$y_T - y_B$};
\draw[black!70] (3.14,0.86) -- (3.14,-0.05) node[below, font=\scriptsize] {$x$};
\node[font=\scriptsize, black!70] at (3.14,2.75) {$dx$};
% limits
\draw[black, dashed] (0.4,0) -- (0.4,3.6);
\draw[black, dashed] (5.8,0) -- (5.8,3.9);
\node[font=\scriptsize, black, below] at (0.4,0) {$a$};
\node[font=\scriptsize, black, below] at (5.8,0) {$b$};
\end{tikzpicture}
$$

When both side boundaries collapse to points, the limits $a$ and $b$ are the
intersection abscissas.

> **Worked example.** Find the area enclosed by $y = x^2$ and $y = 2x - x^2$.
>
> The curves meet where $x^2 = 2x - x^2$, that is $2x^2 - 2x = 0$, so $x = 0$
> and $x = 1$. On $[0, 1]$ the parabola $y_T = 2x - x^2$ lies above
> $y_B = x^2$, so
>
> $$
> A = \int_0^1 \bigl[(2x - x^2) - x^2\bigr]\, \d x
>   = \int_0^1 (2x - 2x^2)\, \d x
>   = \left[ x^2 - \tfrac{2}{3}x^3 \right]_0^1
>   = \frac{1}{3}.
> $$

### When the curves cross

If $f$ and $g$ trade places inside $[a, b]$, the single formula
$A = \int_a^b |f(x) - g(x)|\, \d x$ still measures the geometric area, but the
integral must be split at each crossing so that each piece uses the correct top
curve.

> **Worked example.** Find the area between $y = \sin x$ and $y = \cos x$ on
> $[0, \pi/2]$.
>
> The curves cross where $\sin x = \cos x$, at $x = \pi/4$. Cosine is on top
> for $x < \pi/4$ and sine for $x > \pi/4$, so the integral splits at the
> crossing:
>
> $$
> A = \int_0^{\pi/4} (\cos x - \sin x)\, \d x
>   + \int_{\pi/4}^{\pi/2} (\sin x - \cos x)\, \d x.
> $$
>
> The first piece is $[\sin x + \cos x]_0^{\pi/4} = \sqrt{2} - 1$; the second is
> $[-\cos x - \sin x]_{\pi/4}^{\pi/2} = \sqrt{2} - 1$. Their sum is
> $2\sqrt{2} - 2$.

Skipping the split and integrating $\cos x - \sin x$ straight across would let
the negative second piece cancel the first, reporting a smaller number than the
true area. The absolute value partitions the interval by which curve is higher.

### Integrating with respect to y

Some regions are described more naturally by left and right boundaries
$x = f(y)$ and $x = g(y)$. Then horizontal strips of height $\Delta y$ are the
natural cut, and

$$
A = \int_c^d (x_R - x_L)\, \d y,
$$

with $x_R$ the right boundary and $x_L$ the left.

> **Worked example.** Find the area enclosed by the line $y = x - 1$ and the
> parabola $y^2 = 2x + 6$.
>
> Solving the two equations simultaneously gives the intersection points
> $(-1, -2)$ and $(5, 4)$. Horizontal strips are the natural cut. Solving the
> parabola for $x$, the right boundary is $x_R = y + 1$ and the left is
> $x_L = \tfrac{1}{2}y^2 - 3$, so
>
> $$
> A = \int_{-2}^{4} \left[ (y + 1) - \bigl(\tfrac{1}{2}y^2 - 3\bigr) \right] \d y
>   = \int_{-2}^{4} \left( -\tfrac{1}{2}y^2 + y + 4 \right) \d y = 18.
> $$

Integrating in $x$ here would force the region into two pieces, because the
bottom boundary switches from the lower branch of the parabola to the line. The
choice of variable is a choice of which description needs fewer integrals.

| Description | Strip | Height/width | Formula |
| --- | --- | --- | --- |
| Top/bottom curves $y = f(x)$, $y = g(x)$ | vertical, width $\Delta x$ | $y_T - y_B$ | $\int_a^b (y_T - y_B)\, \d x$ |
| Left/right curves $x = f(y)$, $x = g(y)$ | horizontal, height $\Delta y$ | $x_R - x_L$ | $\int_c^d (x_R - x_L)\, \d y$ |

## Volumes by cross-section

The area idea lifts one dimension. Let $S$ be a solid lying between the planes
$x = a$ and $x = b$, and let $A(x)$ be the area of the cross-section cut by the
plane through $x$ perpendicular to the $x$-axis. Slicing $S$ into slabs of
thickness $\Delta x$, the slab at $x_i^\ast$ is nearly a cylinder of base area
$A(x_i^\ast)$ and height $\Delta x$, with volume $A(x_i^\ast)\, \Delta x$.

> **Definition (Volume by cross-section).** If the cross-sectional area $A(x)$
> is a continuous function of $x$ on $[a, b]$, the volume of the solid is
>
> $$
> V = \lim_{n \to \infty} \sum_{i=1}^n A(x_i^\ast)\, \Delta x = \int_a^b A(x)\, \d x.
> $$

Each volume below is a special choice of the cross-section $A(x)$, set up the
same way.

```algorithm
caption: $\textsc{Volume}$ — setting up a solid as an integral of cross-sections
choose an axis (say $x$) and the slicing planes perpendicular to it
for each slice at position $x$ do
  identify the cross-section (disk, washer, triangle, square, …)
  express its area $A(x)$ in terms of $x$ alone
determine the limits $a, b$ where the solid begins and ends
return $\int_a^b A(x)\, dx$
```

> **Worked example.** Find the volume of a pyramid of height $h$ with a square
> base of side $L$.
>
> Place the apex at the origin with the axis along $x$. Similar triangles make
> the cross-section at position $x$ a square of side $s = Lx/h$, so
> $A(x) = (Lx/h)^2$. The pyramid runs from $x = 0$ to $x = h$:
>
> $$
> V = \int_0^h \frac{L^2}{h^2} x^2\, \d x = \frac{L^2}{h^2} \cdot \frac{h^3}{3}
>   = \frac{L^2 h}{3},
> $$
>
> the classical one-third base times height.

$$
% caption: A pyramid sliced perpendicular to its axis; similar triangles make
% the slice a square of side s that grows linearly from apex O to the base.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.3,0) -- (6.2,0) node[right] {$x$};
% pyramid profile (isoceles triangle), apex O at the origin
\draw[black, thick] (0,0) -- (5.2,1.9) -- (5.2,-1.9) -- cycle;
\node[black, left, font=\scriptsize] at (0,0) {$O$};
% base side L
\draw[<->, black!70] (5.6,1.9) -- (5.6,-1.9);
\node[font=\scriptsize, black!70, right] at (5.6,1.35) {$L$};
% slice of side s at position x
\draw[acc, very thick] (2.9,1.06) -- (2.9,-1.06);
\node[font=\scriptsize, acc, left] at (2.9,0.6) {$s$};
\node[font=\scriptsize, black!70, below right] at (2.9,-1.06) {$x$};
% height h
\draw[<->, black] (0,-2.3) -- (5.2,-2.3) node[midway, below, font=\scriptsize] {$h$};
\end{tikzpicture}
$$

No revolution is involved; the only input is the area of a moving slice.

## Solids of revolution: disks and washers

When the region under $y = f(x)$ is revolved about the $x$-axis, each
cross-section perpendicular to the axis is a **disk** of radius $f(x)$, so
$A(x) = \pi [f(x)]^2$.

$$
% caption: Revolving the region under y = f(x) about the x-axis; the slice at x
% is a solid disk of radius f(x), giving cross-sectional area pi f(x) squared.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.4,0) -- (6.6,0) node[right] {$x$};
% curve
\draw[black, thick] (0.4,0.55) .. controls (2.2,2.3) and (4.0,1.0) .. (6.0,2.0)
  node[right, black] {$y = f(x)$};
% region under the curve
\filldraw[acc!10, draw=none] (0.4,0) -- (0.4,0.55)
  .. controls (2.2,2.3) and (4.0,1.0) .. (6.0,2.0) -- (6.0,0) -- cycle;
% representative disk at x (appears as a vertical ellipse, edge-on)
\filldraw[acc!14, draw=acc, thick] (3.0,0) ellipse (0.24 and 1.32);
\draw[black!70] (3.0,0) -- (3.0,1.32);
\node[font=\scriptsize, black!70, right] at (3.05,0.72) {$f(x)$};
\draw[black!70] (3.0,0) -- (3.0,-0.05) node[below, font=\scriptsize] {$x$};
\node[font=\scriptsize, acc, above] at (3.0,1.5) {disk};
\end{tikzpicture}
$$

> **Worked example.** Revolve the semicircle $y = \sqrt{r^2 - x^2}$ about the
> $x$-axis to recover the volume of a sphere.
>
> Each cross-section is a disk of radius $\sqrt{r^2 - x^2}$, so
> $A(x) = \pi(r^2 - x^2)$ and
>
> $$
> V = \int_{-r}^{r} \pi (r^2 - x^2)\, \d x
>   = 2\pi \left[ r^2 x - \frac{x^3}{3} \right]_0^{r}
>   = \frac{4}{3}\pi r^3.
> $$

If the region lies between two curves, revolving it leaves a hole, and the
cross-section is a **washer** — an outer disk with a smaller disk removed. Its
area is the difference of the two disk areas.

> **Definition (Disk and washer method).** Revolving about a horizontal or
> vertical axis, the cross-section perpendicular to the axis is a disk or washer
> and
>
> $$
> A = \pi (\text{outer radius})^2 - \pi (\text{inner radius})^2,
> $$
>
> with the inner radius $0$ for a solid disk. The volume is $\int A\, \d x$ or
> $\int A\, \d y$ along the axis of revolution.

$$
% caption: Revolving the region between two curves leaves a washer cross-section
% whose area is the outer disk minus the inner disk.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axis
\draw[->, black] (-0.4,0) -- (6.6,0) node[right] {$x$};
% outer and inner radii annotus (drawn as an annulus in end-on view)
\begin{scope}[shift={(1.8,1.5)}]
  \filldraw[acc!12, draw=acc, thick] (0,0) circle (1.5);
  \filldraw[white, draw=black, thick] (0,0) circle (0.62);
  \draw[acc, ->] (0,0) -- (150:1.5);
  \node[acc, font=\scriptsize] at (150:0.95) {$R$};
  \draw[black!70, ->] (0,0) -- (-40:0.62);
  \node[black!70, font=\scriptsize] at (-33:0.4) {$r$};
  \node[font=\scriptsize, black, below] at (0,-1.7) {cross-section};
\end{scope}
% solid of revolution schematic
\begin{scope}[shift={(4.9,0)}]
  \draw[black, thick] (-0.9,1.5) .. controls (0,1.15) and (0.4,1.3) .. (1.0,1.55);
  \draw[black, thick] (-0.9,0.6) .. controls (0,0.95) and (0.4,0.8) .. (1.0,0.55);
  \draw[black!55] (-0.9,1.5) arc (90:-90:0.28 and 0.75);
  \draw[black!55, dashed] (-0.9,1.5) arc (90:270:0.28 and 0.75);
  \draw[black!55] (1.0,1.55) arc (90:-90:0.30 and 0.80);
  \node[font=\scriptsize, black] at (0.05,-0.35) {solid};
\end{scope}
\end{tikzpicture}
$$

> **Worked example.** Revolve the region between $y = x$ and $y = x^2$ about
> the $x$-axis, then about the line $y = 2$.
>
> The curves meet at $(0,0)$ and $(1,1)$; on $[0,1]$ the line is the outer
> boundary and the parabola the inner, so the outer radius is $x$ and the inner
> is $x^2$:
>
> $$
> V = \int_0^1 \pi \bigl[ x^2 - (x^2)^2 \bigr]\, \d x
>   = \pi \int_0^1 (x^2 - x^4)\, \d x
>   = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1
>   = \frac{2\pi}{15}.
> $$
>
> Revolving about $y = 2$ only shifts the radii. The farther curve gives the
> larger radius, so the outer radius is $2 - x^2$ and the inner is $2 - x$:
>
> $$
> V = \pi \int_0^1 \bigl[(2 - x^2)^2 - (2 - x)^2\bigr]\, \d x = \frac{8\pi}{15}.
> $$

## Solids of revolution: cylindrical shells

The washer method is awkward for some revolutions. Rotating the region under
$y = 2x^2 - x^3$ (from $x = 0$ to $x = 2$) about the _y_-axis would require
solving a cubic for $x$ in terms of $y$ to find the washer radii. The **shell
method** avoids the inversion by slicing the region into vertical strips
_parallel_ to the axis of revolution. Each strip sweeps out a thin cylindrical
shell.

A shell of radius $x$, height $f(x)$, and thickness $\Delta x$ has volume
approximately $2\pi x \cdot f(x) \cdot \Delta x$ — circumference times height
times thickness — as seen by cutting the shell along its side and unrolling it
into a flat slab.

$$
% caption: A cylindrical shell of radius x and height f(x), cut and unrolled
% into a slab of dimensions 2 pi x by f(x) by the thickness dx.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% the shell (a tall thin cylinder shown in perspective)
\begin{scope}[shift={(0,0)}]
  \draw[black, thick] (0,0) ellipse (0.95 and 0.32);
  \draw[black, thick] (-0.95,0) -- (-0.95,2.7);
  \draw[black, thick] (0.95,0) -- (0.95,2.7);
  \draw[black, thick] (0,2.7) ellipse (0.95 and 0.32);
  \draw[black!55, dashed] (-0.95,0) arc (180:360:0.95 and 0.32);
  \draw[black!70, ->] (0,0) -- (0.95,0);
  \node[black, font=\scriptsize, below] at (0.5,-0.05) {$x$};
  \draw[<->, black!70] (1.25,0.15) -- (1.25,2.55) node[midway, right, font=\scriptsize] {$f(x)$};
  \node[font=\scriptsize, black, below] at (0,-0.55) {shell};
\end{scope}
% arrow: cut and unroll
\draw[->, black, very thick] (2.4,1.35) -- (3.5,1.35) node[midway, above, font=\scriptsize] {unroll};
% the unrolled slab
\begin{scope}[shift={(3.9,0.15)}]
  \filldraw[acc!12, draw=acc, thick] (0,0) rectangle (3.4,2.7);
  \draw[<->, black!70] (0,-0.28) -- (3.4,-0.28) node[midway, below, font=\scriptsize] {base $=$ circumference};
  \draw[<->, black!70] (3.62,0) -- (3.62,2.7) node[midway, right, font=\scriptsize] {$f(x)$};
\end{scope}
\end{tikzpicture}
$$

> **Definition (Shell method).** Revolving about a vertical axis the region
> under $y = f(x) \ge 0$ from $x = a$ to $x = b$ (with $0 \le a < b$), the volume
> is
>
> $$
> V = \int_a^b 2\pi x\, f(x)\, \d x = \int_a^b (\text{circumference})(\text{height})\, \d x.
> $$
>
> For revolution about a different vertical line, replace the radius $x$ by the
> distance from the strip to the axis.

> **Worked example.** Revolve the region under $y = 2x^2 - x^3$, $0 \le x \le 2$,
> about the $y$-axis.
>
> A vertical strip at $x$ sweeps a shell of radius $x$ and height $f(x)$, so
>
> $$
> V = \int_0^2 2\pi x (2x^2 - x^3)\, \d x
>   = 2\pi \int_0^2 (2x^3 - x^4)\, \d x
>   = 2\pi \left[ \frac{x^4}{2} - \frac{x^5}{5} \right]_0^2
>   = \frac{16\pi}{5}.
> $$
>
> No cubic is solved and no local maximum is located; the strip parallel to the
> axis sidesteps both.

### Choosing between disks and shells

Neither method is universally easier. The deciding question is which variable
describes the region with fewer integrals and a tractable integrand. A strip
_perpendicular_ to the axis of revolution becomes a disk or washer; a strip
_parallel_ to it becomes a shell.

| | Disks / washers | Cylindrical shells |
| --- | --- | --- |
| Strip orientation | perpendicular to axis | parallel to axis |
| Radius given by | the strip's distance out along the slice | the strip's distance to the axis |
| Integrate in | the variable along the axis | the variable across the region |
| Favored when | region set by top/bottom curves, axis horizontal | region set by outer curve, axis vertical, hard inversion |

When both setups are available they must agree, and the choice is only about
which integrand is simpler.

> **Worked example.** Find the volume when the region under $y = \sqrt{x}$,
> $0 \le x \le 1$, is revolved about the $x$-axis — by disks, then by shells.
>
> **Disks** (strip perpendicular to the axis). The radius is $\sqrt{x}$, so
>
> $$
> V = \int_0^1 \pi\bigl(\sqrt{x}\bigr)^2\, \d x = \pi \int_0^1 x\, \d x
>   = \frac{\pi}{2}.
> $$
>
> **Shells** (strip parallel to the axis). Relabel the curve as $x = y^2$. A
> horizontal strip at height $y$ sweeps a shell of radius $y$ and height
> $1 - y^2$, so
>
> $$
> V = \int_0^1 2\pi y\,(1 - y^2)\, \d y
>   = 2\pi \left[ \frac{y^2}{2} - \frac{y^4}{4} \right]_0^1
>   = \frac{\pi}{2}.
> $$
>
> The two agree.

Shells win outright when inversion is the obstacle. For the region between
$y = x$ and $y = x^2$ revolved about the $y$-axis, a strip of radius $x$ and
height $x - x^2$ gives

$$
V = \int_0^1 2\pi x (x - x^2)\, \d x
  = 2\pi \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1
  = \frac{\pi}{6},
$$

whereas the washer method would demand splitting and inverting both curves.

[^area-def]: Stewart, §5.1 — the area formula as the limit of Riemann sums of strip areas $[f(x_i^\ast) - g(x_i^\ast)]\,\Delta x$, and its reduction to the area under one curve when $g \equiv 0$.