Volume of revolution — disk method, washer method

WHY does slicing work?

We cannot find the volume of a curvy solid directly. But we can find the volume of a thin slice, because a thin slice is approximately a cylinder, and we know a cylinder's volume: $V_{\text{cyl}} = (\text{area of circular face}) \times (\text{thickness}) = \pi r^2 \, h.$

If we spin around the x-axis and slice perpendicular to it, each slice has thickness $dx$ and a circular face whose radius is the height of the curve, $r = f(x)$.

Deriving the Disk Method from scratch

WHAT: Region bounded by $y=f(x)$, the x-axis, between $x=a$ and $x=b$, rotated about the x-axis.

HOW: Cut at position $x$, a thin slab of thickness $\Delta x$.

• The slab is (approximately) a cylinder lying on its side.
• Its radius = distance from axis to curve = $f(x)$.
• Its thickness = $\Delta x$.

So the slab volume: $\Delta V \approx \pi\,[f(x)]^2\,\Delta x.$ Why this step? Area of the circular face is $\pi r^2 = \pi[f(x)]^2$; multiply by thickness.

Sum all slabs and let $\Delta x \to 0$ — that limit is the definite integral: $\boxed{\,V = \pi \int_a^b [f(x)]^2 \, dx\,}$

Deriving the Washer Method

WHAT: Now the region is between two curves, $y=f(x)$ (outer, top) and $y=g(x)$ (inner, bottom), rotated about the x-axis. Because the region sits above the axis with a gap, each slice has a hole.

HOW: The slice is a big disk of radius $R = f(x)$ with a small disk of radius $r = g(x)$ removed: $\Delta V \approx \big(\pi R^2 - \pi r^2\big)\Delta x = \pi\big([f(x)]^2 - [g(x)]^2\big)\Delta x.$ Why this step? Area of a ring = (outer circle area) − (inner circle area). Subtract, not square the difference!

$\boxed{\,V = \pi \int_a^b \Big([R(x)]^2 - [r(x)]^2\Big)\, dx\,}$

Worked Example 1 — Disk

Region under $y=\sqrt{x}$, from $x=0$ to $x=4$, rotated about the x-axis.

$V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx.$ Why this step? Radius $=\sqrt{x}$, so $[f(x)]^2 = x$. The square kills the root — clean.

$= \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\cdot\frac{16}{2} = 8\pi.$

Worked Example 2 — Washer

Region between $y=x$ and $y=x^2$ (for $0\le x\le 1$), rotated about the x-axis.

Which is outer? On $(0,1)$, $x > x^2$, so $R=x$ (far from axis), $r=x^2$ (near).

$V = \pi\int_0^1\big(x^2 - (x^2)^2\big)dx = \pi\int_0^1 (x^2 - x^4)\,dx.$ Why this step? $R^2 = x^2$, $r^2 = (x^2)^2 = x^4$. Square each radius separately.

$= \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac13 - \frac15\right) = \pi\cdot\frac{2}{15} = \frac{2\pi}{15}.$

Worked Example 3 — Rotation about a non-axis line $y=2$

Region under $y=x$, $0\le x \le 2$, rotated about $y=2$.

Here the axis is the line $y=2$. Distance from axis to curve $= |x - 2| = 2 - x$ on $[0,2]$. It's a disk (the region touches the axis at $x=2$): $V=\pi\int_0^2 (2-x)^2\,dx.$ Why this step? New radius is measured from $y=2$, not from $y=0$.

Let $u=2-x$: $V=\pi\int_0^2 u^2\,du \cdot(\text{sign handled by symmetry}) = \pi\left[\frac{(x-2)^3}{3}\right]_0^2 = \pi\left(0 - \frac{(-2)^3}{3}\right)=\frac{8\pi}{3}.$

Common Mistakes

Active Recall

Connections

• Definite Integral as a Limit of Sums — slicing → Riemann sum → integral.
• Area Between Curves — same setup, but integrand differs ($f-g$ vs $R^2-r^2$).
• Shell Method — alternative for revolution, slices parallel to axis.
• Volume by Cross-Sections — disk/washer are special cases (circular cross-sections).
• u-substitution — used to evaluate shifted-axis integrals.

Concept Map

Hinglish (regional understanding)