Integration adds up infinitely many infinitely thin things. For
area the thin
things were strips; for volume they are slices. Cut a solid into wafer-thin
slabs perpendicular to the x-axis. If the slab at position
x has cross-sectional area A(x) and thickness
dx, its volume is A(x)\,dx. Stack the slabs
from a to b and the total volume is
V = \int_a^{b} A(x)\,dx.
The art is finding A(x). The most common case — and the one we derive
below — is a solid of revolution, where every cross-section is a circle.
The disk method, derived slab by slab
Take the region under y = f(x) on [a, b] and
spin it about the x-axis. It sweeps out a solid; each cross-section is a
filled circle — a disk.
Step 1 — find the radius of one slice. At position x,
the curve sits at height f(x). When that point whirls around the axis it
traces a circle of radius exactly r = f(x).
Step 2 — the cross-sectional area is a circle's. A disk of radius
f(x) has area
A(x) = \pi r^2 = \pi\, f(x)^2.
Step 3 — one thin disk's volume. Give the slab thickness
\Delta x; it is a squat cylinder of volume
\Delta V = \pi\, f(x)^2\,\Delta x.
Step 4 — stack the disks and refine. Sum over a partition and let the slices get
infinitely thin; the Riemann sum becomes an integral:
V = \lim_{\Delta x \to 0} \sum \pi\, f(x_i)^2\,\Delta x = \pi \int_a^{b} f(x)^2\,dx.
Worked example: rotate y = \sqrt{x} on [0, 1]
Step 1 — square the radius. Here f(x) = \sqrt{x}, so
f(x)^2 = x — the messy square root vanishes the moment it is squared.
Step 2 — assemble the integral.
V = \pi \int_0^{1} \big(\sqrt{x}\big)^2 dx = \pi \int_0^{1} x\,dx.
Step 3 — integrate and evaluate.
V = \pi \left[\frac{x^2}{2}\right]_0^{1} = \pi \cdot \frac{1}{2} = \frac{\pi}{2}.
The trumpet-shaped solid swept out by y = \sqrt{x} holds exactly
\pi/2 cubic units.
A solid of revolution about the x-axis, with cross-sections
perpendicular to that axis, has volume V = \int_a^{b} A(x)\,dx, where:
-
Disk — rotating the region under
y = f(x) gives circular cross-sections of area
A(x) = \pi f(x)^2, so
\displaystyle V = \pi \int_a^{b} f(x)^2\,dx.
-
Washer — rotating the region between an outer
R(x) and an inner r(x) leaves an
annular (ring-shaped) cross-section of area
A(x) = \pi\big(R(x)^2 - r(x)^2\big), so
\displaystyle V = \pi \int_a^{b} \big(R(x)^2 - r(x)^2\big)\,dx — the
hole is subtracted as its own disk.
The formula V = \int A(x)\,dx never assumed a circle — that was a
special case. Some of the prettiest solids have cross-sections that are squares, triangles, or
semicircles, with no rotation in sight.
Take a solid whose base is the disk x^2 + y^2 \le 1 and whose slices
perpendicular to the x-axis are squares. At position
x the base runs from y = -\sqrt{1 - x^2} to
+\sqrt{1 - x^2}, a side of length
s = 2\sqrt{1 - x^2}. The square slice has area
A(x) = s^2 = 4(1 - x^2), so
V = \int_{-1}^{1} 4(1 - x^2)\,dx = 4\left[x - \frac{x^3}{3}\right]_{-1}^{1} = 4 \cdot \frac{4}{3} = \frac{16}{3}.
No \pi anywhere — the moment the cross-section stops being a circle,
the \pi simply doesn't appear. Find A(x)
honestly and the integral does the rest.