Areas Between Curves and Volumes

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 $f (x_{i}^{*}) Δ x$ ; sum the pieces; take the limit. The sum becomes $\int_{a}^{b} f (x) d x$ , evaluated by the Fundamental Theorem. Area under a single curve is the first instance; 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 $f (x) \geq g (x)$ throughout. Slicing $S$ into vertical strips of width $Δ x$ , the strip at $x_{i}^{*}$ is nearly a rectangle of height $f (x_{i}^{*}) - g (x_{i}^{*})$ . Summing the strip areas and passing to the limit gives a Riemann sum whose value is the integral of the gap.

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.

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.

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

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.

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 $Δ y$ are the natural cut, and

A = c \int d (x_{R} - x_{L}) d y,

with $x_{R}$ the right boundary and $x_{L}$ the left.

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 $Δ x$	$y_{T} - y_{B}$	$\int_{a}^{b} (y_{T} - y_{B}) d x$
Left/right curves $x = f (y)$ , $x = g (y)$	horizontal, height $Δ y$	$x_{R} - x_{L}$	$\int_{c}^{d} (x_{R} - x_{L}) d y$

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 $Δ x$ , the slab at $x_{i}^{*}$ is nearly a cylinder of base area $A (x_{i}^{*})$ and height $Δ x$ , with volume $A (x_{i}^{*}) Δ x$ .

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

Algorithm:

\textsc{Volume}

— setting up a solid as an integral of cross-sections

1
choose an axis (say $x$ ) and the slicing planes perpendicular to it
2
for each slice at position $x$ do
3
identify the cross-section (disk, washer, triangle, square, …)
4
express its area $A(x)$ in terms of $x$ alone
5
determine the limits $a, b$ where the solid begins and ends
6
return $\int_a^b A(x)\, dx$

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.

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 $A (x) = π [f (x)]^{2}$ .

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.

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.

Revolving the region between two curves leaves a washer cross-section whose area is the outer disk minus the inner disk.

Solids of revolution: cylindrical shells

The washer method is awkward for some revolutions. Rotating the region under $y = 2 x^{2} - x^{3}$ (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 $Δ x$ has volume approximately $2 π x \cdot f (x) \cdot Δ x$ — circumference times height times thickness — as seen by cutting the shell along its side and unrolling it into a flat slab.

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.

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.

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 = 0 \int 1 2 π x (x - x^{2}) d x = 2 π [\frac{x ^{3}}{3} - \frac{x ^{4}}{4}]_{0}^{1} = \frac{π}{6},

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

Stewart, §5.1 — the area formula as the limit of Riemann sums of strip areas $[f (x_{i}^{*}) - g (x_{i}^{*})] Δ x$ , and its reduction to the area under one curve when $g \equiv 0$ . ↩