Which volume-of-revolution method do I use?
Disk, washer, shell, or cross-sections — read the axis and the shape before picking.
First ask what the cross-sections perpendicular to the axis of revolution look like. A solid circle (no hole) means the disk method. A circle with a hole down the middle (revolving a region that doesn't touch the axis) means the washer method. If solving for x in terms of y (or vice versa) would be awkward, switch to the shell method instead, which integrates along the OTHER variable. If the solid isn't a revolution at all — just a stack of known cross-sectional shapes — use general cross-sectional slicing directly.
Disk and washer — cross-sections perpendicular to the axis
Formula
Disk method
If the region under y=f(x) on [a,b] (with f(x)\ge 0) is revolved about the x-axis, the resulting solid's volume is V = \pi\int_a^b \big[f(x)\big]^2\,dx.
Example
Show proof
This is the cross-sectional slicing formula V=\int_a^b A(x)\,dx applied to a specific shape: revolving the segment from the axis up to f(x) sweeps out a full circle of radius f(x) at each x, so the cross-section there is a disk with area A(x) = \pi\big[f(x)\big]^2. Substituting that area function directly into the slicing formula gives V = \pi\int_a^b [f(x)]^2\,dx; no new idea is needed beyond identifying the disk's area correctly.
Formula
Washer method
If the region between y=f(x) (outer) and y=g(x) (inner), with f(x)\ge g(x)\ge 0 on [a,b], is revolved about the x-axis, the resulting solid's volume is V = \pi\int_a^b \Big(\big[f(x)\big]^2 - \big[g(x)\big]^2\Big)\,dx.
Outer radius squared minus inner radius squared — never (outer minus inner) squared.
Show proof
Revolving the outer boundary alone would sweep out a solid disk of area \pi[f(x)]^2 at each x, by the disk method. But the inner boundary carves a circular hole of area \pi[g(x)]^2 out of that disk at the same x, leaving a washer — an annulus — of area \pi[f(x)]^2 - \pi[g(x)]^2. Substituting that washer's area into the cross-sectional slicing formula V=\int_a^b A(x)\,dx gives the stated integral.
Section
The Disk Method
Section
The Washer Method
Shells — when solving for the other variable is awkward
Formula
Shell method
If the region under y=f(x) on [a,b] (with 0\le a, f(x)\ge 0) is revolved about the y-axis, the resulting solid's volume is V = 2\pi\int_a^b x\,f(x)\,dx.
Example
Show proof
Instead of slicing perpendicular to the axis of revolution, slice parallel to it: a thin vertical strip at position x, of width dx and height f(x), sweeps out a thin cylindrical shell when revolved about the y-axis. That shell has radius x, height f(x), and thickness dx, so unrolling it into a flat rectangular slab gives a volume (\text{circumference})\times(\text{height})\times(\text{thickness}) = (2\pi x)\,f(x)\,dx. Summing infinitely many such shells from x=a to x=b — the same limiting Riemann-sum argument behind every slicing formula — turns the sum into V = 2\pi\int_a^b x f(x)\,dx.
Section
Cylindrical Shells
Section
Choosing a Volume Method
Not a revolution at all
If a solid lies between x=a and x=b, and the cross-sectional area of the solid taken perpendicular to the x-axis at position x is A(x), then the solid's volume is V = \int_a^b A(x)\,dx.
Use directly whenever the solid's slices are given (squares, semicircles, triangles) rather than generated by spinning a region.
Show proof
Slice the solid into n thin slabs, each of thickness \Delta x. The i-th slab is nearly a cylinder of cross-sectional area A(x_i^*) and height \Delta x, so its volume is approximately A(x_i^*)\Delta x. Summing over all n slabs gives a Riemann sum \sum_{i=1}^n A(x_i^*)\Delta x that approximates the total volume, and the approximation only improves as the slabs get thinner. Taking the limit as n\to\infty turns the Riemann sum into the definite integral \int_a^b A(x)\,dx by the same limiting argument that defines any definite integral.
Section