1. The unifying picture: slice, approximate, integrate
Every application in this topic — and most of integration in physics and engineering — follows the same four-step pattern. Internalize it once and you stop memorizing formulas; you re-derive them.
- Slice. Cut the quantity into many thin pieces, each one indexed by a variable $x$ (or $y$, or $\theta$) running over $[a, b]$. The piece at position $x$ has width $dx$.
- Approximate. Over that thin slice, pretend the quantity is constant or simple — a rectangle, a thin disk, a tiny straight segment. Write down its contribution as some function $\phi(x)\,dx$.
- Sum. Add up the contributions of all the slices. That's a Riemann sum, $\sum \phi(x_i)\,\Delta x$.
- Integrate. Take the limit as the slices get infinitely thin: $\sum \to \int_a^b \phi(x)\,dx$.
An integral is the limit of a Riemann sum. Any "total" quantity that can be approximated by slicing into thin pieces can be expressed as an integral.
The only thing that changes from one application to the next is what $\phi(x)\,dx$ is — the geometric or physical content of one thin slice. Area: a strip of height $f(x) - g(x)$ and width $dx$. Volume by disk: a coin of radius $f(x)$ and thickness $dx$. Work: a force $F(x)$ acting over a distance $dx$. Same machine, different ingredients.
If you only learn the formulas as a list — "area = $\int (\text{top} - \text{bottom})$, volume = $\pi \int f^2$, work = $\int F \, dx$" — you'll forget them by next semester and you won't recognize the pattern when a new application appears (electric charge, hydrostatic pressure, expected value, the flux of a vector field). Learn the slicing recipe instead. Then every formula is a one-line derivation away.
2. Area between two curves
Suppose $f(x) \geq g(x)$ on $[a, b]$. The region trapped between the two curves and the vertical lines $x = a$ and $x = b$ has area
$$ A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx. $$Apply the slicing recipe to see why: take a vertical strip at position $x$, infinitely thin (width $dx$), running from the lower curve up to the upper. Its height is $f(x) - g(x)$, so its area is $[f(x) - g(x)]\,dx$. Integrate.
Finding the limits
Often $a$ and $b$ aren't given directly — you have to find them as the $x$-values where the curves cross. Set $f(x) = g(x)$ and solve. If you get more than two intersection points, the curves swap which is on top, and you must split the integral at each crossing, taking the absolute value (or, equivalently, putting "top minus bottom" on each piece).
Sometimes a region is much simpler to describe as $g(y) \leq x \leq f(y)$ for $y \in [c, d]$ — think of a region bounded by $x = y^2$ and $x = y$. Then slice horizontally: the strip at height $y$ has width $f(y) - g(y)$, and $A = \int_c^d [f(y) - g(y)]\,dy$. Pick the variable that lets you write the bounding curves as single-valued functions; you'll save a lot of grief.
3. Volumes of revolution: disk and washer
Take a region in the plane and spin it 360° around a line. The result is a three-dimensional solid of revolution. The classic example: spin the region under $y = \sqrt{x}$ on $[0, 4]$ about the $x$-axis and you get a paraboloid shaped like a horn.
To find the volume, slice the solid perpendicular to the axis of revolution. Each slice is a flat coin — a disk, if the region touches the axis, or a washer (a disk with a hole), if there's a gap.
Disk method
The region under $y = f(x) \geq 0$ on $[a, b]$, rotated about the $x$-axis, has volume
$$ V = \pi \int_a^b \bigl[f(x)\bigr]^2 \, dx. $$Each thin coin at position $x$ has radius $f(x)$ and thickness $dx$, so its volume is $\pi[f(x)]^2\,dx$.
That's it — the $\pi r^2$ for the area of the coin's face, times $dx$ for its thickness, integrated. The slicing recipe again.
Washer method
If the region doesn't touch the axis — say it's the area between $y = f(x)$ (outer) and $y = g(x)$ (inner), with $f \geq g \geq 0$ — each slice is a washer with outer radius $f(x)$ and inner radius $g(x)$. The face area is $\pi[f(x)]^2 - \pi[g(x)]^2$, so
$$ V = \pi \int_a^b \bigl(\,[f(x)]^2 - [g(x)]^2\,\bigr)\,dx. $$Crucially, this is not $\pi \int [f - g]^2$ — you have to subtract the squares, not square the difference. The hole is a separate disk; it doesn't combine into the outer one algebraically.
4. Volumes of revolution: shell method
The disk method slices the solid perpendicular to the axis. The shell method slices it parallel to the axis — into thin cylindrical tubes, like nested layers of an onion.
Take a region in the plane under $y = f(x) \geq 0$ on $[a, b]$ (with $a \geq 0$) and rotate it about the $y$-axis. Pick a thin vertical strip at position $x$, of height $f(x)$ and width $dx$. When spun around the $y$-axis, that strip sweeps out a cylindrical shell of radius $x$, height $f(x)$, and thickness $dx$.
The lateral surface area of a cylinder is $2\pi r \cdot h$ (circumference times height), and a thin shell's volume is that area times its thickness:
$$ dV = \underbrace{2\pi x}_{\text{circumference}} \cdot \underbrace{f(x)}_{\text{height}} \cdot \underbrace{dx}_{\text{thickness}}. $$Integrate:
$$ V = 2\pi \int_a^b x \, f(x)\, dx. $$5. Disk vs shell: when each is easier
Disk and shell can always solve each other's problems — they just take different routes. The right method is the one that gives you the easier integral.
The two examples below describe the same solid: the region under $y = x^2$ on $[0, 2]$, rotated about the $y$-axis. Watch how the choice of method changes the bookkeeping.
The same solid, two ways
Method A — shell (slice parallel to the $y$-axis)
At position $x \in [0, 2]$, take a vertical strip of height $f(x) = x^2$ and width $dx$. Rotated about the $y$-axis, this strip becomes a shell of radius $x$ and height $x^2$:
$$ V = 2\pi \int_0^2 x \cdot x^2 \,dx = 2\pi \int_0^2 x^3\,dx = 2\pi \cdot \frac{2^4}{4} = 8\pi. $$One integral, one term, done in three lines. Shell is the natural choice because we already have $y$ as a function of $x$, and the axis of revolution is the $y$-axis.
Method B — disk (slice perpendicular to the $y$-axis)
Now we slice horizontally. At height $y$, the slice is a disk whose radius runs from the $y$-axis out to the curve. We need $x$ as a function of $y$: $y = x^2 \implies x = \sqrt{y}$. The disk at height $y$ has radius $\sqrt y$, and $y$ runs from $0$ to $4$ (since $x = 2 \implies y = 4$). But the region we're spinning is bounded on the right by $x = 2$ as well, so the disks are actually washers from $\sqrt y$ out to $2$:
$$ V = \pi \int_0^4 \bigl(2^2 - (\sqrt y)^2\bigr)\,dy = \pi \int_0^4 (4 - y)\,dy = \pi\bigl[4y - \tfrac{y^2}{2}\bigr]_0^4 = \pi(16 - 8) = 8\pi. $$Same answer, more steps: we had to invert the function, change variables, and set up a washer (not a plain disk) to handle the outer boundary.
Match the slicing direction to the variable you already have. If the curve is given as $y = f(x)$ and you're rotating about the $y$-axis, shells are easier. If rotating about the $x$-axis, disks are easier. Reverse the picture (curves as $x = g(y)$, rotation about the $x$-axis) and shells come back. The rule of thumb: you want the slice variable to match the axis the function is "naturally" written in terms of, so you don't have to invert.
Summary table
| Method | Slice direction | Formula | Natural setup |
|---|---|---|---|
| Disk | ⊥ to axis | $\pi \int [f(x)]^2\,dx$ | $y = f(x)$, rotate about $x$-axis |
| Washer | ⊥ to axis | $\pi \int ([R]^2 - [r]^2)\,dx$ | annular region; rotate about an axis the region doesn't touch |
| Shell | ∥ to axis | $2\pi \int x \, f(x)\,dx$ | $y = f(x)$, rotate about $y$-axis |
6. Arc length
How long is a curve $y = f(x)$ between $x = a$ and $x = b$? Slice it into tiny pieces — each piece is, to first order, a little straight segment with horizontal extent $dx$ and vertical extent $dy = f'(x)\,dx$. By Pythagoras, its length is
$$ ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + [f'(x)]^2}\,dx. $$Add up all the tiny straight pieces:
$$ L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx. $$This is again the slicing recipe — each piece is approximated by a straight line, and the integral is the limit. The factor $\sqrt{1 + (f')^2}$ shows up everywhere a length-along-the-curve element appears (surface area, work along a curved path, line integrals).
Arc-length integrals are notorious for being analytically intractable. Even something as innocent as $y = x^2$ gives $L = \int \sqrt{1 + 4x^2}\,dx$, which requires a trigonometric or hyperbolic substitution. Anything more complicated than a polynomial of low degree typically has to be evaluated numerically. The formula is exact; the closed form is rare.
7. Surface area of revolution
Spin the curve $y = f(x) \geq 0$ on $[a, b]$ about the $x$-axis. The resulting surface (think of the outside of a vase) has area
$$ SA = 2\pi \int_a^b f(x) \sqrt{1 + [f'(x)]^2}\,dx. $$The slicing argument: a tiny piece of curve of length $ds$ at height $f(x)$, swept around the axis, traces out a thin band — essentially a frustum, but in the limit a cylinder of radius $f(x)$ and width $ds$. Its lateral area is $2\pi f(x)\,ds$, and substituting $ds = \sqrt{1 + (f')^2}\,dx$ gives the formula.
Note the structure: it's just $2\pi r \cdot ds$, integrated. Circumference times arc-length element. If you ever forget the formula, re-derive it from that mental picture.
8. Average value of a function
What does it mean to take the "average" of a function $f$ on $[a, b]$ — not the average of finitely many sample values, but the average of all its values over the interval? The answer follows the same slicing logic.
The average of $n$ samples $f(x_1), \dots, f(x_n)$ is $\frac{1}{n}\sum f(x_i)$. If we sample uniformly with spacing $\Delta x = (b-a)/n$, multiply numerator and denominator by $\Delta x$:
$$ \frac{1}{n}\sum f(x_i) = \frac{1}{n \Delta x}\sum f(x_i)\,\Delta x = \frac{1}{b - a}\sum f(x_i)\,\Delta x. $$Let $n \to \infty$; the Riemann sum becomes an integral:
Geometrically: $\bar f$ is the height of the rectangle on $[a, b]$ that has the same area as the region under $y = f(x)$.
That rectangle picture is the mean value theorem for integrals: if $f$ is continuous on $[a, b]$, then there exists some $c \in [a, b]$ with $f(c) = \bar f$. The continuous function actually achieves its own average somewhere on the interval.
Concrete example: the average of $f(x) = x^2$ on $[0, 3]$ is
$$ \bar f = \frac{1}{3} \int_0^3 x^2\,dx = \frac{1}{3} \cdot \frac{27}{3} = 3. $$So the parabola $y = x^2$, between $x = 0$ and $x = 3$, takes the value $3$ on average — and by the MVT for integrals, it equals $3$ at $x = \sqrt 3$.
9. Work: a physical application
Push a block along a straight line with a constant force $F$ over a distance $d$ and the work you do is $W = F \cdot d$. But what if the force depends on position — like a spring that pushes back harder the more you stretch it?
The same recipe rescues us. Slice the path $[a, b]$ into tiny intervals of width $dx$. On each interval, the force is approximately constant at $F(x)$, so the work done over that thin piece is $F(x)\,dx$. Sum and integrate:
$$ W = \int_a^b F(x)\,dx. $$This is the integral version of "work = force times distance," exactly analogous to how an integral generalizes a sum of constant pieces.
Hooke's law and the spring
A spring stretched a distance $x$ from its natural length pulls back with force $F(x) = kx$ (Hooke's law), where $k$ is the spring constant. The work to stretch it from $0$ to $L$ is
$$ W = \int_0^L kx\,dx = \frac{kL^2}{2}. $$This is the elastic potential energy stored in the spring at extension $L$ — the same $\tfrac{1}{2}kL^2$ that appears in introductory physics. The integral isn't just bookkeeping; the antiderivative is a physical quantity.
Hydrostatic force on a dam, the gravitational work to pump water out of a tank, the energy needed to lift a chain, the kinetic energy of a rotating disk, the electric charge in a region of varying density — every one of these problems is a slice-and-integrate problem. Identify the thin slice. Write down its contribution as $\phi(x)\,dx$. Integrate.
10. Common pitfalls
When computing area between two curves, blindly writing $\int [f - g]$ without checking which is larger gives the wrong sign — and if the curves swap order on the interval, the right answer is the sum of pieces, not one integral.
Disk and washer formulas carry a $\pi$; shell carries $2\pi$. A common slip is to write the integral correctly but drop the constant in front. Sanity-check your answer's units (volume should have $\pi$ in it for a solid of revolution) before moving on.
The washer formula is $\pi\int([f]^2 - [g]^2)\,dx$, not $\pi\int (f - g)^2\,dx$. Squaring the difference is a different number entirely. The two disks (outer and inner) don't combine algebraically inside the square — only inside the integral.
Set up the slicing direction by looking at the axis of revolution and how the function is given. If you find yourself inverting $y = f(x)$ to write $x = f^{-1}(y)$ just so you can use disks, stop — shells were the right call.
$\sqrt{1 + [f'(x)]^2}$ almost never gives a tidy antiderivative. Don't panic when the integral resists technique; reach for numerical methods (Simpson's rule, $\texttt{scipy.integrate.quad}$) when the exact form isn't required.
Work is in joules — newtons times metres. If $F(x)$ is in newtons and $x$ is in metres, $\int F\,dx$ is automatically in joules. Mixing centimetres and metres, or pounds and kilograms, silently corrupts the answer. Keep units consistent before integrating.
11. Worked examples
Try each one before opening the solution. The goal is to recognize the slicing pattern — not to memorize formulas.
Example 1 · Area between $y = x$ and $y = x^2$ on $[0, 1]$
Step 1. Find which is larger. On $[0, 1]$ we have $x \geq x^2$ (since $x^2 = x \cdot x \leq x$ when $0 \leq x \leq 1$). So $y = x$ is the top curve.
Step 2. Set up the integral:
$$ A = \int_0^1 (x - x^2)\,dx. $$Step 3. Evaluate:
$$ A = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}. $$Example 2 · Volume of the solid: $y = \sqrt x$ rotated about the $x$-axis on $[0, 4]$ (disk method)
Step 1. The region touches the axis (the lower bound is $y = 0$), so plain disks — no washers.
Step 2. Disk of radius $f(x) = \sqrt x$ at position $x$:
$$ V = \pi \int_0^4 (\sqrt x)^2\,dx = \pi \int_0^4 x\,dx. $$Step 3. Evaluate:
$$ V = \pi \cdot \frac{x^2}{2}\bigg|_0^4 = \pi \cdot 8 = 8\pi. $$Example 3 · Volume of the same solid as Example 2 — but via the shell method
The region under $y = \sqrt x$ on $[0, 4]$, rotated about the $x$-axis. To use shells we slice parallel to the $x$-axis — i.e., horizontally — so each shell has radius $y$ (its distance from the $x$-axis).
Step 1. Find the horizontal range of $y$: as $x$ runs from $0$ to $4$, $y = \sqrt x$ runs from $0$ to $2$.
Step 2. At height $y$, the strip runs from $x = y^2$ (the curve, rewritten) out to $x = 4$. So its width is $4 - y^2$. The shell has radius $y$, height $4 - y^2$, thickness $dy$:
$$ V = 2\pi \int_0^2 y\,(4 - y^2)\,dy = 2\pi \int_0^2 (4y - y^3)\,dy. $$Step 3. Evaluate:
$$ V = 2\pi\left[2y^2 - \frac{y^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi. $$Same answer as the disk method — but notice that disk was a one-line setup ($\pi \int x\,dx$), while shell needed an inverted function and a width $4 - y^2$. For this rotation, disk is plainly easier. Different geometries flip the verdict; that's the point of having both tools.
Example 4 · Arc length of $y = x^{3/2}$ on $[0, 4]$
Step 1. Compute $f'(x) = \tfrac{3}{2}x^{1/2}$, so $[f'(x)]^2 = \tfrac{9x}{4}$.
Step 2. Set up the arc-length integral:
$$ L = \int_0^4 \sqrt{1 + \tfrac{9x}{4}}\,dx. $$Step 3. Substitute $u = 1 + \tfrac{9x}{4}$, $du = \tfrac{9}{4}dx$, so $dx = \tfrac{4}{9}\,du$. When $x = 0$, $u = 1$; when $x = 4$, $u = 10$:
$$ L = \frac{4}{9}\int_1^{10} \sqrt u\,du = \frac{4}{9}\cdot\frac{2}{3}\bigl[u^{3/2}\bigr]_1^{10} = \frac{8}{27}\bigl(10\sqrt{10} - 1\bigr). $$Numerically, $L \approx 9.07$. Note how rare it is for this to work out cleanly — the choice $y = x^{3/2}$ was carefully arranged so that $1 + (f')^2$ is a polynomial in $x$.
Example 5 · Average value of $f(x) = \sin x$ on $[0, \pi]$
Step 1. Apply the definition:
$$ \bar f = \frac{1}{\pi - 0}\int_0^\pi \sin x\,dx = \frac{1}{\pi}\bigl[-\cos x\bigr]_0^\pi. $$Step 2. Evaluate the antiderivative:
$$ \bar f = \frac{1}{\pi}(-\cos\pi + \cos 0) = \frac{1}{\pi}(1 + 1) = \frac{2}{\pi} \approx 0.637. $$So on its first half-arch, the sine function takes the value $2/\pi$ on average. By the MVT for integrals, $\sin c = 2/\pi$ for some $c \in (0, \pi)$ — in fact for two such $c$ (by symmetry).
Example 6 · Work to stretch a spring from rest to $0.4$ m, with $k = 80\,\text{N/m}$
Step 1. By Hooke's law, $F(x) = 80x$ newtons.
Step 2. Work:
$$ W = \int_0^{0.4} 80x\,dx = 80 \cdot \frac{x^2}{2}\bigg|_0^{0.4} = 40 \cdot (0.16) = 6.4\;\text{J}. $$Equivalently, $\tfrac{1}{2}kL^2 = \tfrac{1}{2}(80)(0.4)^2 = 6.4\,\text{J}$ — confirming the closed form $W = \tfrac{1}{2}kL^2$.