Recipe

Find the volume of a solid of revolution

Use when: a flat region is revolved around an axis, and you need the resulting solid's volume

Goal: a single number -- the exact volume, found by integrating cross-sectional area rather than measuring the solid directly

Taught in Volumes by disks and washers

1. 1

   Draw the flat region being revolved, and mark exactly which line it spins around.

   Why: the axis determines what "radius" even means for this problem -- the same region revolved around a different axis produces a different solid entirely.

2. 2

   Check whether the region's bottom boundary IS the axis of revolution.

   • the region touches the axis along its whole bottom edge — use the disk method, with R(x)=\text{top}(x).
   • the region is bounded away from the axis on both sides — use the washer method, with outer radius R(x)=\text{top}(x) and inner radius r(x)=\text{bottom}(x).

   Why: a disk is really just a washer whose inner radius happens to be 0 -- the two methods are one idea, split only by whether there's a hole.

3. 3

   For a disk, the area is \pi[R(x)]^2. For a washer, it's \pi\big([R(x)]^2-[r(x)]^2\big).

   Example

   Revolving the region between y=\sqrt x (top) and y=x^2 (bottom) on [0,1]: R(x)=\sqrt x, r(x)=x^2, so the area is \pi\big[(\sqrt x)^2-(x^2)^2\big]=\pi(x-x^4).

   Why: squaring R(x) and r(x) SEPARATELY, before subtracting, is essential. An annulus's area genuinely depends on the two radii's squares taken one at a time, not on the square of their difference.

4. 4

   Set up V=\pi\int_a^b [R(x)]^2\,dx (disk) or V=\pi\int_a^b\big([R(x)]^2-[r(x)]^2\big)\,dx (washer), using the region's own bounds as a and b.

   Example

   V=\pi\int_0^1(x-x^4)\,dx=\pi\left[\dfrac{x^2}{2}-\dfrac{x^5}{5}\right]_0^1=\pi\left(\dfrac12-\dfrac15\right)=\dfrac{3\pi}{10}.

   Why: this is the same "slice and sum" logic every earlier accumulation in this course used — a solid's volume is just the running total of its cross-sectional areas.