← Unit 8

Solids of Revolution

Take a flat region, spin it around an axis, and it sweeps out a 3D solid. The whole topic is one idea — slice the solid into discs — with a few wrinkles in how you measure the radius.

1 · Spin a region into a solid

Here's the region under y = √x on [0, 4]. Drag the sweep slider to revolve it around the x-axis and watch the solid form. Then flip on "show discs" — the solid is really just a stack of thin discs, and that's the whole secret.

drag to orbit · scroll to zoom
V = —
Sweep angle:
Slide the disc  x = 2.50
SweepRotates the region around the x-axis. Stop partway to see the solid half-formed; the badge shows the volume so far.
Show discsSwaps the smooth solid for the stack of discs it's really made of — the picture behind the integral.
Slice + RHighlights one disc (amber) and its radius R. Slide it along to watch R = √x grow and shrink.

Why does that give a clean formula? Look at one disc. Its radius is the height of the region, R = √x. A disc is just a circle with thickness, so its area is πR² — and stacking all of them is an integral:

one representative slice
radius
disc area
volume

▶ Now build one from your curves

The model above is one example. Here's the same machine — but you type the curves, straight from your homework. Pick the axis (x, y, or any line like y=3), hit Revolve, and it hands you the disc / washer / shell setup in your own numbers — then press Sweep to watch the volume build.

2 · It's all one formula

Most guides hand you a "disc formula" and a separate "washer formula" and make you memorize which to use. Skip all that. There is one formula for every solid of revolution, and disc is just a special case of it.

R is the outer radius (axis to the far edge of the region); r is the inner radius (axis to the near edge). That's it — every problem is this integral.

So when is it a disc versus a washer? It's the same question as: does the region touch the axis? If it touches, there's no hole — the inner radius is r = 0, and π(R²−0²) collapses to plain πR². A disc is nothing but a washer whose hole has shrunk to nothing.

Touches the axis
r = 0 → disc
Gap to the axis
r > 0 → washer

Stop thinking "which method?" Always write π∫(R²−r²). The only real work — the entire skill — is measuring R and r. That's next.

3 · The whole skill: finding the radius

If one formula covers everything, then every problem comes down to one thing: get R and r right. Three principles handle every case the AP exam can throw at you.

1 · The radius is a distance, not the curve

The radius is how far the edge of the region sits from the axis — a distance you measure, not simply "the function." They happen to be equal only when the axis is the x-axis itself (distance from y = 0 is just the height). Move the axis and that shortcut breaks. Always ask "distance from the axis to the edge," and you'll never get tripped up.

2 · The axis decides your variable

The radius always points straight out from the axis — perpendicular to it. That one fact tells you whether to integrate in x or y, which is the step most students guess at:

Axis of revolutionRadius pointsBuilt fromIntegrate
x-axis, or any y = kvertically ↕y-values: curve − axisdx
y-axis, or any x = khorizontally ↔x-values: curve − axisdy

Horizontal axis → vertical radius → it's made of y's → dx. Vertical axis → horizontal radius → x's → dy (so solve the curves for x first). No guessing.

3 · Distance = far − near (so shift with the axis)

Every radius is one edge minus another: outer R = far edge − axis, inner r = near edge − axis. When the axis is the x-axis, "− axis" is "− 0" and disappears. When it isn't, you shift by the axis — the step that costs the most points:

about y = 0 (x-axis)
about y = −2 (below the region)
about y = 5 (above the region)

Same curve y = √x every time — only the axis moved, so only the "− axis" piece changed. Set it up as far − near and the radius always comes out positive.

4 · The three cases, worked

Watch the one formula handle all three. Notice each is the same recipe from above — only how you measure the radius changes.

Disc — region touches the axis (r = 0)

Problem. Revolve the region under y = √x on [0, 4] about the x-axis. Find the volume.
1
Radius (a distance). Distance from the x-axis up to the curve is √x. It touches the axis, so r = 0.
2
One formula. π(R²−r²) with r = 0 is just πR² — the √ disappears when squared.
3
Integrate (horizontal axis → dx) over the bounds.
Volume =

Washer — a gap leaves a hole (r > 0)

Want to see why that formula is an integral? Step through the whole build on one washer — find its radii, sum the slices, take the limit, land on the integral.

The plan, in 4 steps:  ① one washer's area  →  ② add them all up  →  ③ take the limit  →  ④ that's the integral
drag to orbit
Problem. Revolve the region between y = x and y = x² about the x-axis. Find the volume.
1
Two distances. The curves meet at x = 0 and x = 1; on that interval the line y = x is farther from the axis, so it's the outer radius and y = x² is the inner.
2
Far − near. Outer R = x, inner r = x².
3
One formula, then integrate. π(R² − r²).
Volume = 2π/15

Shifted axis — the radius shifts too

Problem. Revolve the region under y = √x on [0, 4] about the line y = −2. Find the volume.
1
Outer radius = far − axis. Distance from y = −2 up to the curve: √x − (−2).
2
Inner radius = near − axis. The region's bottom (y = 0) is still 2 above the axis, so r = 2. The gap makes it a washer.
3
One formula, then integrate.
Volume = 88π/3

5 · Drill it

Disc or washer? What's the radius? Where's the axis? Six questions on the decisions that actually get tested.

Score: 0 / 0
That's the last big method of Unit 8. Up next is the BC-only Arc Length — or jump to the Exam Practice capstone. Need a quick reference? The cheat sheet has every formula on one page.