Chapter 18 — Applications of Integration

18.1 One Pattern, Eight Problems

The Fundamental Theorem of Calculus (Chapter 14) taught you that a definite integral is the limit of a Riemann sum and that you can evaluate it by antidifferentiation. But the integral computes far more than signed area. Every quantity that can be built up as an accumulation over a single parameter is a definite integral in disguise: volume, length, surface area, mass, work, force, the location of a balance point.

This chapter develops the eight most important of those applications. They look unrelated — what does the area between two parabolas have to do with the work needed to drain a tank? — but they share one skeleton, and once you see the skeleton you can build new applications you have never been taught.

The skeleton has four steps, and we will name them the slice-and-sum recipe:

1. Draw the picture and pick a slicing variable ($x$, $y$, or $t$).
2. Cut a thin slice of the quantity at a general position, of thickness $dx$ (or $dy$, $dt$).
3. Express that slice's contribution — call it $dA$, $dV$, $dW$, $dM$ — as a function of the slicing variable times the thickness.
4. Integrate over the full range of the variable.

The Key Insight. Setting up the integral is the hard part; computing it is the easy part. Steps 1–3 are where all the thinking lives — choosing the right slice and writing its contribution correctly. Step 4 is mechanical, drawing on the techniques of Chapters 15–16, or on a computer. Every section of this chapter is really a lesson in modeling: turning a geometric or physical picture into a single integrand. Master the picture and the calculus follows.

This is theme 6 of the book — approximation is the soul of calculus — wearing a geometric costume. We approximate a curved region by thin straight-edged slivers, sum the slivers, and let their thickness shrink to zero. It is also theme 2: geometry and algebra are inseparable. Every formula below has a picture, and we will never write the formula without first drawing the picture.

18.2 Area Between Curves

Start with the application closest to the definite integral itself. Suppose two continuous functions satisfy $f(x) \geq g(x)$ on $[a, b]$. The region trapped between their graphs has area

$$A = \int_a^b \big[\,f(x) - g(x)\,\big] \, dx.$$

Slice and sum. Cut a thin vertical strip at position $x$. Its top sits on $y = f(x)$, its bottom on $y = g(x)$, so its height is $f(x) - g(x)$ and its width is $dx$. The strip is essentially a rectangle of area $[f(x) - g(x)]\,dx$. Sum the strips across $[a,b]$ — that is the integral.

Geometric Intuition. Picture sweeping a vertical broom from $x=a$ to $x=b$. At each instant the broom has length equal to the vertical gap between the two curves. The area is the gap accumulated across the sweep. Notice that the formula does not care whether the curves are above or below the $x$-axis — only the gap $f-g$ matters, because the parts of each curve below the axis contribute identical negative pieces that cancel in the subtraction. This is why "top minus bottom" always works, even when both curves dip negative.

Worked Example 18.2.1 — The standard case

Find the area between $y = x^2$ and $y = 2x$.

Setup first. Sketch the upward parabola $y=x^2$ and the line $y=2x$. They meet where $x^2 = 2x$, i.e. $x(x-2)=0$, so at $x=0$ and $x=2$. Between these, test $x=1$: the line gives $2$, the parabola gives $1$, so the line is on top. The slice has height $2x - x^2$.

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

Worked Example 18.2.2 — When the curves cross

Find the total area enclosed between $y = \sin x$ and $y = \cos x$ on $[0, 2\pi]$.

Setup first. On $[0,2\pi]$ the two curves cross at $x = \pi/4$ and $x = 5\pi/4$ (where $\tan x = 1$). The "top" curve switches at each crossing, so we cannot use a single integrand. Check a test point in each piece: on $[0,\pi/4]$, $\cos \geq \sin$; on $[\pi/4, 5\pi/4]$, $\sin \geq \cos$; on $[5\pi/4, 2\pi]$, $\cos \geq \sin$. Always integrate larger minus smaller:

$$A = \int_0^{\pi/4}(\cos x - \sin x)\,dx + \int_{\pi/4}^{5\pi/4}(\sin x - \cos x)\,dx + \int_{5\pi/4}^{2\pi}(\cos x - \sin x)\,dx.$$

Each antiderivative is elementary; carrying out the arithmetic gives $\sqrt2 + 2\sqrt2 + \sqrt2 = 4\sqrt2 \approx 5.657$.

Common Pitfall. Many students compute area between crossing curves with a single integral $\int_a^b (f-g)\,dx$ and get a wrong, sometimes negative, answer. The bare integral computes signed area: where $g > f$, the integrand goes negative and cancels positive area elsewhere. Area is always non-negative. When the curves cross inside $[a,b]$, you must split at every crossing and integrate $|f-g|$ piece by piece — or, equivalently, integrate "top minus bottom" on each subinterval. Find the crossings before you integrate.

Horizontal strips: integrating in $y$

Sometimes a region is described most naturally by its left and right boundaries, $x = f_L(y)$ and $x = f_R(y)$, with $f_R(y) \geq f_L(y)$ on $[c,d]$. Then slice horizontally: a strip at height $y$ has width $f_R(y) - f_L(y)$ and thickness $dy$, so

$$A = \int_c^d \big[\,f_R(y) - f_L(y)\,\big]\, dy.$$

When to switch. Use horizontal strips whenever a vertical strip would change which curve forms its top or bottom partway across the region — a region bounded on the left by $x = y^2$ and on the right by $x = y+2$, for instance, needs only one horizontal integral but two vertical ones. Choosing the slicing variable to minimize splitting is the central setup decision of this whole chapter.

Check Your Understanding. Set up (do not evaluate) the area of the region bounded by $x = y^2$ and $x = y + 2$.

Answer

The curves meet where $y^2 = y+2$, i.e. $y^2 - y - 2 = (y-2)(y+1) = 0$, so $y = -1$ and $y = 2$. For $y$ in $(-1,2)$, the line $x = y+2$ lies to the right of the parabola $x = y^2$. Horizontal strips give the single clean integral $A = \int_{-1}^{2}\big[(y+2) - y^2\big]\,dy$. (Vertical strips would require splitting at $x=1$ because the parabola supplies both the top and bottom boundary for $x<1$.) Evaluating: $\big[\tfrac{y^2}{2} + 2y - \tfrac{y^3}{3}\big]_{-1}^{2} = \tfrac{9}{2}$.

18.3 Volumes by Disks and Washers

A solid of revolution is the solid swept out when a planar region is rotated about an axis. A cone, a sphere, a wine barrel, a lathe-turned table leg — all are solids of revolution. We compute their volume by slicing perpendicular to the axis of rotation; each slice is a flat circular disk whose volume is easy.

The disk method

Take the region under $y = f(x) \geq 0$ on $[a,b]$ and rotate it about the $x$-axis. Slice perpendicular to the axis, at position $x$, with thickness $dx$. The slice is a thin disk (a coin) of radius $f(x)$ and thickness $dx$, so its volume is $dV = \pi[f(x)]^2\,dx$. Sum:

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

Geometric Intuition. Imagine the solid sitting on a deli slicer aligned with the $x$-axis. Each cut produces a circular coin whose radius is the height of the original curve at that $x$. The coin's face area is $\pi r^2 = \pi f(x)^2$; multiply by the coin's thickness $dx$ to get its volume; stack the coins from $a$ to $b$. The radius of the coin is the distance from the axis to the curve — that single phrase generalizes to every volume formula in this section.

Worked Example 18.3.1 — The cone, derived

Rotate the line $y = \dfrac{r}{h}\,x$ on $[0, h]$ about the $x$-axis. At $x=h$ the height is $r$, so this sweeps out a cone of base radius $r$ and height $h$. The disk at position $x$ has radius $f(x) = \tfrac{r}{h}x$:

$$V = \int_0^h \pi \left(\frac{r}{h}\,x\right)^2 dx = \frac{\pi r^2}{h^2}\int_0^h x^2\,dx = \frac{\pi r^2}{h^2}\cdot \frac{h^3}{3} = \frac{1}{3}\pi r^2 h.$$

The familiar cone formula $V = \tfrac13 \pi r^2 h$ falls out of one integral — no clever decomposition, no limit of stacked frustums by hand.

Worked Example 18.3.2 — The sphere, derived

Rotate the upper semicircle $y = \sqrt{R^2 - x^2}$ on $[-R, R]$ about the $x$-axis. The disk at $x$ has radius $\sqrt{R^2-x^2}$, so $[f(x)]^2 = R^2 - x^2$:

$$V = \int_{-R}^{R} \pi (R^2 - x^2)\,dx = \pi\left[R^2 x - \frac{x^3}{3}\right]_{-R}^{R} = 2\pi\left(R^3 - \frac{R^3}{3}\right) = \frac{4}{3}\pi R^3.$$

Historical Note. Archimedes (c. 287–212 BC) found the volume of the sphere by an ingenious balancing argument, comparing slices of a sphere, a cone, and a cylinder on an imaginary lever. He was so proud of the result $V_{\text{sphere}} = \tfrac{4}{3}\pi R^3$ — and of the fact that the sphere is exactly two-thirds of its circumscribing cylinder — that he asked for a sphere-in-cylinder to be carved on his tombstone. Cicero rediscovered that overgrown tomb 137 years later. What cost Archimedes a lifetime of genius is now a two-line calculus exercise; that is the power FTC put in your hands.

The washer method

If the region lies between two curves $y = f(x)$ (outer) and $y = g(x)$ (inner), with $f \geq g \geq 0$, then rotating about the $x$-axis carves a hole through each disk. The slice becomes a washer — a disk with a circular bite removed. Its area is the big circle minus the small circle, $\pi f(x)^2 - \pi g(x)^2$, so

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

Common Pitfall. The washer integrand is $f^2 - g^2$, not $(f-g)^2$. The area of a washer is (outer area) minus (inner area) $= \pi R^2 - \pi r^2$, and you cannot factor that into $\pi(R-r)^2$. Squaring the gap $f-g$ is the single most common volume error. Square each radius, then subtract.

Worked Example 18.3.3 — A washer

Rotate the region between $y = \sqrt{x}$ (outer) and $y = x$ (inner) on $[0,1]$ about the $x$-axis.

Setup first. On $(0,1)$, $\sqrt{x} > x$, so $\sqrt x$ is the outer radius and $x$ the inner radius. The washer at position $x$ has area $\pi(\sqrt x)^2 - \pi x^2 = \pi(x - x^2)$:

$$V = \pi\int_0^1 (x - x^2)\,dx = \pi\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \pi\left(\frac12 - \frac13\right) = \frac{\pi}{6}.$$

18.4 Volumes by Cylindrical Shells

Disks and washers slice perpendicular to the axis of rotation. Sometimes that forces you into ugly algebra — especially when you rotate about a vertical axis but your region is described by $y = f(x)$. To rotate the region under $y = f(x)$ about the $y$-axis with disks, you would first have to invert $f$ to write $x$ as a function of $y$, which may be impossible. The cylindrical-shell method slices parallel to the axis instead and avoids the inversion entirely.

Take the region under $y = f(x) \geq 0$ on $[a,b]$ with $0 \le a$, and rotate it about the $y$-axis. Cut a thin vertical strip at position $x$, thickness $dx$. As it revolves, the strip sweeps out a thin cylindrical shell — a tin can with no top or bottom — of:

• radius $x$ (the distance from the strip to the axis),
• height $f(x)$,
• thickness $dx$.

Unroll the shell into a flat sheet: its length is the circumference $2\pi x$, its height is $f(x)$, its thickness is $dx$. So $dV = 2\pi x\, f(x)\, dx$, and

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

Geometric Intuition. A shell is a soup-can label. Slit the can vertically and flatten it: you get a rectangle of width $2\pi x$ (the circumference it used to wrap around), height $f(x)$, and thickness $dx$. Disks stack like coins along the axis; shells nest like Russian dolls around the axis. When the curve is given as $y=f(x)$ and you spin about a vertical line, shells keep you in your original variable — that is the whole reason they exist.

Worked Example 18.4.1 — A basic shell

Rotate the region under $y = x^2$ on $[0,1]$ about the $y$-axis.

$$V = \int_0^1 2\pi x \cdot x^2 \, dx = 2\pi \int_0^1 x^3 \, dx = 2\pi \cdot \frac14 = \frac{\pi}{2}.$$

To do this with washers you would invert to $x = \sqrt y$ and integrate in $y$ — doable here, but shells are cleaner, and for a curve like $y = x^2 + x$ the inversion is genuinely unpleasant.

Worked Example 18.4.2 — Rotating about a shifted axis

Rotate the region under $y = x^2$ on $[0,1]$ about the vertical line $x = 2$.

Setup first. The only thing that changes is the radius. A strip at position $x$ now sits a distance $2 - x$ from the axis $x=2$ (and since $0 \le x \le 1 < 2$, this distance is positive). The shell radius is $2 - x$:

$$V = \int_0^1 2\pi (2 - x)\, x^2 \, dx = 2\pi \int_0^1 (2x^2 - x^3)\,dx = 2\pi\left[\frac{2}{3} - \frac{1}{4}\right] = 2\pi\cdot\frac{5}{12} = \frac{5\pi}{6}.$$

The Key Insight. Disk/washer versus shell is a convenience choice, not a correctness one — both give the same volume. Choose by asking: is my slice perpendicular or parallel to the axis cleaner? Slice perpendicular (disks/washers) when the region is described as a function of the same variable as the axis; slice parallel (shells) when it is described as a function of the other variable. The radius in either method is always "distance from the slice to the axis," and a shifted axis only changes that distance.

Check Your Understanding. You want the volume when the region under $y = e^{-x^2}$ on $[0,2]$ is rotated about the $y$-axis. Which method, and why?

Answer

Shells. The disk/washer method would require solving $y = e^{-x^2}$ for $x$ as a function of $y$ — possible ($x = \sqrt{-\ln y}$) but messy, and it splits awkwardly. Shells keep the original variable: $V = \int_0^2 2\pi x\, e^{-x^2}\,dx$. As a bonus this one is doable by hand via the substitution $u = x^2$, giving $\pi(1 - e^{-4})$. The integrand $x e^{-x^2}$ has an elementary antiderivative while $e^{-x^2}$ alone does not — the shell's extra factor of $x$ rescues us.

18.5 Arc Length

How long is a curve? Stretch a thread along $y = f(x)$ from $x=a$ to $x=b$ and measure the thread. The slice-and-sum recipe handles it: approximate the curve by many short straight segments and add their lengths.

A short segment over a horizontal step $dx$ rises by $dy = f'(x)\,dx$. By the Pythagorean theorem its length is

$$ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\;dx = \sqrt{1 + [f'(x)]^2}\;dx.$$

This $ds$ is the arc length element — remember it; it reappears in surface area below and in line integrals in Chapter 35. Summing,

$$L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx.$$

Geometric Intuition. Zoom in on the curve until a tiny piece looks straight. That piece is the hypotenuse of a right triangle with legs $dx$ (horizontal) and $dy = f'(x)\,dx$ (vertical). The slope $f'(x)$ tilts the hypotenuse; the steeper the curve, the more the factor $\sqrt{1+(f')^2}$ exceeds $1$, and the more arc length you accumulate per unit of horizontal travel. On a flat stretch ($f'=0$) the factor is $1$ and arc length equals horizontal length, exactly as it should.

Worked Example 18.5.1 — A curve that works out

Find the arc length of $y = x^{3/2}$ on $[0,1]$.

Here $f'(x) = \tfrac{3}{2}x^{1/2}$, so $1 + [f'(x)]^2 = 1 + \tfrac{9}{4}x$. Then

$$L = \int_0^1 \sqrt{1 + \tfrac{9}{4}x}\,dx.$$

Substitute $u = 1 + \tfrac94 x$, $du = \tfrac94\,dx$, so $dx = \tfrac49\,du$; limits run $u: 1 \to \tfrac{13}{4}$:

$$L = \frac{4}{9}\int_1^{13/4}\sqrt{u}\,du = \frac{4}{9}\cdot\frac{2}{3}\,u^{3/2}\Big|_1^{13/4} = \frac{8}{27}\left[\left(\frac{13}{4}\right)^{3/2} - 1\right] \approx 1.440.$$

Why arc length is usually impossible by hand

Example 18.5.1 was rigged: $x^{3/2}$ makes $1+(f')^2$ a perfect setup for substitution. Most arc-length integrals have no elementary antiderivative. Even the humble parabola $y = x^2$ leads to $\int \sqrt{1 + 4x^2}\,dx$, which needs trigonometric substitution (Chapter 16) and produces a formula with a logarithm in it.

Warning. The perimeter of an ellipse is the most famous arc length with no elementary formula. Setting it up gives the complete elliptic integral of the second kind, a function that provably cannot be written with polynomials, roots, exponentials, logs, and trig (Liouville's theorem, Chapter 14, §14.12). So the circle's circumference is a clean $2\pi R$, but squash the circle the tiniest bit and its perimeter escapes elementary functions entirely. This is not a gap in your technique — it is a theorem about what formulas can express. When the antiderivative does not exist in closed form, you integrate numerically; §18.10 shows how.

Parametric curves

If a curve is given parametrically as $(x(t), y(t))$ for $t \in [a,b]$, the same Pythagorean slice gives $ds = \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$, so

$$L = \int_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt.$$

We treat parametric curves in full in Chapter 25; this is a preview of how naturally arc length generalizes once you stop insisting the curve be a graph $y=f(x)$.

18.6 Surface Area of Revolution

Rotating a curve (not a region) about an axis sweeps out a surface — the outside of a vase, the skin of a sphere. To find its area, reuse the arc-length element $ds$.

Rotate $y = f(x) \geq 0$ about the $x$-axis. A short arc of length $ds$ sits at height $f(x)$ above the axis, so when it revolves it traces a thin band (a frustum strip) of circumference $2\pi f(x)$. The band's area is circumference times slant width, $2\pi f(x)\,ds$. With $ds = \sqrt{1 + [f'(x)]^2}\,dx$,

$$S = \int_a^b 2\pi f(x)\,\sqrt{1 + [f'(x)]^2}\,dx.$$

Common Pitfall. Surface area uses the slant arc-length element $ds = \sqrt{1+(f')^2}\,dx$, not the flat width $dx$. Writing $S = \int 2\pi f(x)\,dx$ (forgetting the radical) computes the area of a stack of flat rings, not a slanted surface, and undercounts whenever the curve is tilted. The radical is the difference between wrapping a cone in paper (correct) and stacking washers (wrong).

Worked Example 18.6.1 — The sphere's surface

Rotate $y = \sqrt{R^2 - x^2}$ on $[-R, R]$ about the $x$-axis to sweep out a sphere of radius $R$.

Compute the radical: $f'(x) = \dfrac{-x}{\sqrt{R^2-x^2}}$, so $1 + [f'(x)]^2 = 1 + \dfrac{x^2}{R^2-x^2} = \dfrac{R^2}{R^2-x^2}$, giving $\sqrt{1+[f'(x)]^2} = \dfrac{R}{\sqrt{R^2-x^2}}$. Now watch the radius and the radical cancel beautifully:

$$S = \int_{-R}^{R} 2\pi \sqrt{R^2-x^2}\cdot\frac{R}{\sqrt{R^2-x^2}}\,dx = 2\pi R \int_{-R}^{R} dx = 2\pi R \cdot 2R = 4\pi R^2.$$

The classical surface-area formula $S = 4\pi R^2$, and the cancellation exposes a hidden fact: a sphere's surface area is distributed uniformly along the axis — every slab of equal thickness $dx$ contributes equal surface area, $2\pi R\,dx$. That is Archimedes' hat-box theorem, and it is why a globe's lines of latitude carve equal-area zones for equal north–south spacing.

18.7 Work

In physics, work is force applied through a distance. When the force is constant, $W = F \cdot d$. But forces in the real world vary with position — a spring fights harder the more you stretch it, a tank of water gets heavier to lift from the bottom. The slice-and-sum recipe handles the variation: over an infinitesimal displacement $dx$ the force is essentially constant, so $dW = F(x)\,dx$, and

$$W = \int_a^b F(x)\,dx.$$

This is theme 1 — calculus is the mathematics of change — applied to a force that changes as you move.

Worked Example 18.7.1 — Stretching a spring (Hooke's law)

Hooke's law says a spring resists stretching with restoring force $-kx$, where $x$ is the displacement from rest and $k$ is the stiffness. To stretch it you must apply the opposing force $+kx$. The work to pull it from rest out to extension $L$:

$$W = \int_0^L kx\,dx = \frac{1}{2}kL^2.$$

The energy stored grows with the square of the stretch — the signature of a quadratic potential well. For a spring with $k = 100\ \text{N/m}$ pulled $L = 0.1\ \text{m}$: $W = \tfrac12(100)(0.1)^2 = 0.5\ \text{J}$.

Worked Example 18.7.2 — Pumping water out of a tank

A cylindrical tank of radius $R$ and height $H$ stands full of water. How much work pumps all the water out over the top rim?

Setup first — this is the part that matters. Different layers of water travel different distances, so we cannot use $W = F\cdot d$ with one $d$. Slice the water into thin horizontal slabs. Let $y$ measure height from the bottom of the tank, $0 \le y \le H$. The slab at height $y$ is a coin of volume $\pi R^2\,dy$, mass $\rho\,\pi R^2\,dy$, and weight $\rho g\,\pi R^2\,dy$. To clear the rim it must rise a distance $H - y$. Its work contribution is

$$dW = \underbrace{\rho g\,\pi R^2\,dy}_{\text{weight of slab}}\cdot\underbrace{(H - y)}_{\text{lift distance}}.$$

Integrate over all slabs:

$$W = \int_0^H \rho g\,\pi R^2 (H - y)\,dy = \rho g\,\pi R^2 \left[Hy - \frac{y^2}{2}\right]_0^H = \rho g\,\pi R^2 \cdot \frac{H^2}{2}.$$

For water ($\rho = 1000\ \text{kg/m}^3$, $g = 9.81\ \text{m/s}^2$) in a tank with $R = 2\ \text{m}$, $H = 5\ \text{m}$:

$$W = 1000\cdot 9.81\cdot \pi \cdot 4 \cdot \frac{25}{2} \approx 1.54\times 10^6\ \text{J} \approx 1.5\ \text{MJ},$$

about $0.43\ \text{kWh}$ — a few cents of electricity, but a real load for a pump.

Common Pitfall. In pumping problems, two distances are in play and students mix them up: the slab's position (where the water starts, measured by your variable) and the lift distance (how far that slab must rise to the exit). If $y$ is measured from the bottom and water exits the top at height $H$, the lift distance is $H - y$, not $y$. Always re-read the geometry: define your variable's zero point explicitly, then write the lift distance as "exit height minus slab height." A sign error here silently corrupts the whole integral.

Worked Example 18.7.3 — Lifting a hanging cable

A $50$-ft cable weighing $5\ \text{lb/ft}$ hangs off the edge of a roof. How much work raises the entire cable to the roof?

Setup first. Let $x$ be how far the bottom end has been lifted, $0 \le x \le 50$. At that moment the still-hanging portion has length $50 - x$ and weight $5(50 - x)$ lb. Raising it a further $dx$ costs $5(50-x)\,dx$ ft-lb. (The part already pulled onto the roof is no longer being raised, so it contributes nothing further.)

$$W = \int_0^{50} 5(50 - x)\,dx = 5\left[50x - \frac{x^2}{2}\right]_0^{50} = 5\,(2500 - 1250) = 6250\ \text{ft-lb}.$$

A neat sanity check: the cable's total weight is $250$ lb and its center of mass starts $25$ ft below the roof, so the work should be $250 \times 25 = 6250$ ft-lb. It matches — and that center-of-mass shortcut is exactly the idea of §18.9.

18.8 Hydrostatic Force

A submerged surface — a dam, a porthole, an aquarium wall — feels fluid pressure pushing on it. Pressure grows with depth: at depth $h$ below the surface, a fluid of density $\rho$ exerts pressure $P = \rho g h$. Because pressure varies with depth, the total force is again a slice-and-sum.

Slice the submerged plate into thin horizontal strips. A strip at depth $h$ of width $w(h)$ and height $dh$ has area $w(h)\,dh$. Pressure is essentially constant across so thin a strip, so the force on it is pressure times area:

$$dF = \underbrace{\rho g h}_{\text{pressure}}\cdot\underbrace{w(h)\,dh}_{\text{strip area}}, \qquad F = \int_a^b \rho g h\,w(h)\,dh.$$

Warning. Force is not pressure times total area. The pressure $\rho g h$ is different on every strip, so you cannot pull it out of the integral as a constant. (Only the average pressure times area works, and the average pressure is itself the value of the integral divided by area — which is circular unless you do the integral.) The width $w(h)$ also varies for non-rectangular plates: a triangular dam, a circular porthole, and a trapezoidal trough each have their own $w(h)$ that you must read off the geometry before integrating.

Worked Example 18.8.1 — Force on a vertical dam

A rectangular dam face is $100\ \text{m}$ wide and $50\ \text{m}$ tall, holding back water that reaches the top edge. Find the total hydrostatic force.

Setup first. Measure depth $h$ from the water surface, $0 \le h \le 50$. The dam is rectangular, so the strip width is constant, $w(h) = 100$:

$$F = \int_0^{50}\rho g\,h\cdot 100\,dh = (1000)(9.81)(100)\int_0^{50} h\,dh = 981{,}000\cdot\frac{50^2}{2} = 1.23\times 10^9\ \text{N}.$$

About $1.2\ \text{GN}$ — the weight of roughly $1.2\times 10^8\ \text{kg}$. That staggering number, concentrated toward the base where $h$ is largest, is why real dams are thick at the bottom and taper toward the top.

Real-World Application — Civil engineering: dam and levee design (physics/engineering). The hydrostatic integral is the starting point for every dam, lock gate, and flood levee on Earth. Engineers integrate $\rho g h\,w(h)$ over the wetted face to get total thrust, then take a moment of that force (a second integral, weighting each strip by its depth) to locate the line of action and check the structure against overturning. The triangular cross-section of a gravity dam is a direct consequence: force grows linearly with depth, so the structure must grow proportionally heavier toward its base to resist it.

18.9 Center of Mass and Centroids

Where does an object balance? The center of mass is the single point at which you could support the object and have it stay level — the weighted-average position of its mass. Slice-and-sum delivers it, with one new ingredient: the moment, which weights each slice by its distance from a reference point.

A thin rod (1D)

For a rod along $[a,b]$ with linear density $\rho(x)$ (mass per unit length):

$$M = \int_a^b \rho(x)\,dx \quad(\text{total mass}), \qquad M_O = \int_a^b x\,\rho(x)\,dx \quad(\text{moment about the origin}),$$

$$\bar{x} = \frac{M_O}{M} = \frac{\int_a^b x\,\rho(x)\,dx}{\int_a^b \rho(x)\,dx}.$$

Each slice contributes mass $\rho(x)\,dx$ located at position $x$; the moment is position times mass, summed; the center of mass is total moment divided by total mass. It is exactly the formula for a weighted average, with the integral standing in for the sum.

Geometric Intuition. Picture children on a seesaw. A heavy child near the pivot can balance a light child far away because what matters is mass times distance — the moment. The center of mass is the pivot point where all the moments cancel: $\int (x - \bar x)\,\rho(x)\,dx = 0$. Rearranging that balance condition gives $\bar x = \int x\rho\,dx / \int \rho\,dx$. The integral is just an infinite seesaw with infinitely many infinitesimal children.

Worked Example 18.9.1 — A variable-density rod

A rod on $[0,2]$ has density $\rho(x) = 1 + x$ — light on the left, heavy on the right.

$$M = \int_0^2 (1+x)\,dx = \left[x + \frac{x^2}{2}\right]_0^2 = 2 + 2 = 4,$$ $$M_O = \int_0^2 x(1+x)\,dx = \int_0^2 (x + x^2)\,dx = \left[\frac{x^2}{2} + \frac{x^3}{3}\right]_0^2 = 2 + \frac{8}{3} = \frac{14}{3},$$ $$\bar{x} = \frac{14/3}{4} = \frac{7}{6} \approx 1.17.$$

The balance point sits right of the geometric midpoint $x=1$, exactly as the heavier right end demands. Sanity check passed by intuition.

A flat plate (2D centroid)

For a region of uniform density bounded above by $y = f(x)$ and below by $y = g(x)$ on $[a,b]$, the density cancels and the balance point is the centroid $(\bar x, \bar y)$, a purely geometric quantity:

$$\bar{x} = \frac{1}{A}\int_a^b x\,[f(x)-g(x)]\,dx, \qquad \bar{y} = \frac{1}{A}\int_a^b \frac{1}{2}\big[f(x)^2 - g(x)^2\big]\,dx,$$

where $A$ is the region's area. The $\bar x$ formula weights each vertical strip by its position $x$; the $\bar y$ formula uses the height $\tfrac12(f+g)$ of each strip's own midpoint, which expands to the $\tfrac12(f^2-g^2)$ shown.

Real-World Application — Aerospace and naval architecture (engineering). The centroid of a cross-section and the related centroid of area govern where a beam bends and where a ship floats level. A wing's spar is designed so its neutral axis passes through the centroid of the cross-section, because bending stress is zero there; a ship's stability is computed from the centroid of its submerged volume (the "center of buoyancy") relative to its center of mass. Get the centroid integral wrong and the structure buckles or the ship capsizes — these integrals are not academic.

Math Major Sidebar — Pappus's Centroid Theorem. There is a startling shortcut linking §§18.3–18.4 and §18.9. Pappus's theorem states: if a plane region of area $A$ is revolved about an external axis in its plane, the volume swept out equals $V = 2\pi\bar{d}\,A$, where $\bar d$ is the distance from the centroid to the axis. The centroid travels a circle of circumference $2\pi\bar d$, and "drags" the whole area through it. For instance, a circle of radius $r$ whose center is distance $\,R\,$ from an axis sweeps out a torus; Pappus gives its volume instantly as $V = 2\pi R\cdot \pi r^2 = 2\pi^2 R r^2$ — no shell integral required. Pappus (c. 300 AD) discovered this thirteen centuries before calculus existed; FTC reveals why it is true, since $\int x\,dA$ is both the volume integrand (shells) and $\bar x A$ (centroid).

The 2D and 3D generalization

When the density genuinely varies across a two-dimensional lamina, $\sigma(x,y)$, the single integral is no longer enough; mass becomes a double integral $M = \iint_R \sigma\,dA$ and the moments follow suit. We build the machinery of double and triple integrals in Chapter 32, where volumes of fully three-dimensional solids (not just solids of revolution) and centers of mass of arbitrary bodies become routine. Everything in this chapter is the one-variable warm-up for that.

18.10 Computation: Visualizing a Solid of Revolution

Hand computation builds the understanding; the machine builds the picture and verifies the answer (theme 4). Here we both draw the solid swept out in Worked Example 18.3.1 — the cone from $y = \tfrac{r}{h}x$ — and confirm its volume numerically against the formula $\tfrac13\pi r^2 h$.

# Visualize the solid of revolution from y = (r/h) x rotated about the x-axis,
# and verify its volume V = (1/3) pi r^2 h numerically.
import numpy as np
from scipy.integrate import quad
import matplotlib.pyplot as plt

r, h = 1.0, 2.0
f = lambda x: (r / h) * x                      # the generating curve

# --- numerical check of the disk-method volume ---
V_num, _ = quad(lambda x: np.pi * f(x)**2, 0, h)   # ∫ π f(x)^2 dx
V_exact = (1/3) * np.pi * r**2 * h
print(f"disk-method (quad): {V_num:.6f}")          # 2.094395
print(f"formula 1/3 pi r^2 h: {V_exact:.6f}")      # 2.094395  -> they agree

# --- build the surface of revolution for plotting ---
x = np.linspace(0, h, 60)
theta = np.linspace(0, 2*np.pi, 60)
X, T = np.meshgrid(x, theta)
R = f(X)                                            # radius at each x
Y, Z = R*np.cos(T), R*np.sin(T)                     # spin the curve around the x-axis

ax = plt.figure().add_subplot(projection='3d')
ax.plot_surface(X, Y, Z, alpha=0.7, cmap='viridis')
ax.set_title("Cone as a solid of revolution")
ax.set_xlabel("x"); ax.set_ylabel("y"); ax.set_zlabel("z")
# plt.show()

The 3D plot shows the line $y = \tfrac{r}{h}x$ swept into the familiar cone, and the printed volumes match to six digits — the disk integral and the closed-form formula shaking hands.

Computational Note. scipy.integrate.quad evaluates the disk integral $\int_0^h \pi f(x)^2\,dx$ numerically, without ever using the antiderivative. That it lands on the exact $\tfrac13\pi r^2 h$ is the same theme-4 handshake you saw in Chapter 14: hand calculus gives the exact answer, the numerical routine confirms it, and the 3D surface makes the slicing visible. For arc lengths and surfaces with no elementary antiderivative (the ellipse perimeter of §18.5), quad is not a luxury — it is the only way to get a number at all.

18.11 The Universal Pattern

Step back and notice that every section followed the same four steps — only the slice changed.

The recipe never changes: draw the picture, cut a slice, write the slice's contribution, integrate. Once you trust it, you can attack accumulations you were never explicitly taught — total revenue from a marginal-revenue curve, total charge from a charge density, expected value from a probability density — because they are all the same four steps.

Add to Your Modeling Portfolio. Add an accumulated geometric or physical quantity to your model, set up with the slice-and-sum recipe. State your variable, draw the slice, write its contribution, and integrate. Biology: model the volume of a tumor or organ as a solid of revolution from an imaging profile $r(x)$, $V = \int \pi r(x)^2\,dx$; or the total drug cleared as $\int r(t)\,dt$. Economics: compute consumer surplus as the area between the demand curve and the market price, $\int_0^{q^*}[D(q) - p^*]\,dq$ — an area-between-curves integral with an economic meaning. Physics: compute the work to lift a payload against a variable force (rocket fuel burn, or pumping a non-cylindrical tank), $W = \int F(x)\,dx$, then find the center of mass of a variable-density rod or beam. Data Science: treat a probability density $p(x)$ as a "mass distribution" and compute its mean (center of mass) $\mu = \int x\,p(x)\,dx$ and the area under the curve $\int_a^b p(x)\,dx$ as a probability — the §18.9 moment formula is literally the definition of expected value.

18.12 Summary and Looking Ahead

Integration computes any quantity that accumulates over one parameter. The art is never the algebra — it is the setup: choosing the slicing variable, drawing the slice, and writing its contribution as a differential. Get the slice right and the integral writes itself.

You met eight applications, but they were one idea in eight costumes. Areas and volumes came from geometric slices; arc length and surface area from the Pythagorean arc element $ds$; work, force, and center of mass from physical slices weighted by distance, pressure, or position. The recurring move — approximate the curved by the thin-and-straight, then take the limit — is the soul of calculus (theme 6), and the marriage of every picture to a formula is its inseparable geometry and algebra (theme 2).

Two threads continue beyond this chapter. Chapter 25 generalizes arc length to parametric curves, and Chapter 26 to polar coordinates. Chapter 32 lifts the whole slice-and-sum machine into two and three dimensions with double and triple integrals, where volumes of arbitrary solids — not just solids of revolution — and centers of mass of arbitrary bodies become routine. The single integral you mastered here is the foundation they all stand on.

Chapter 19 closes Part III with the most consequential application of integration of all: differential equations, where the rate of change of a quantity is specified and integration recovers the quantity itself. It is where calculus stops describing static shapes and starts predicting the dynamic world — populations, epidemics, cooling coffee, orbiting planets. The SIR epidemic model (the book's biology anchor) gets its full development there. Turn the page.

Continue to: Exercises · Quiz · Case Study 1 · Case Study 2 · Key Takeaways · Further Reading