The concept is easy: stack up slices to build a solid. The part that actually trips people up is
remembering each shape's area formula — and knowing when to use it. So this page is built as a cheat sheet you
can actually keep.
1 · The method: stack the slices
Take a flat 2D region and slice it into thin strips. On each strip, stand up a known shape —
a square, a triangle, a semicircle. Each shape is a cross section, and its area depends on the strip's
length. Add up all those areas across the region and you get the solid's volume.
That "add up the areas" is exactly an integral. If each cross section sits at position
x and has area A(x), the volume is:
The whole game is writing A(x) correctly — and that has just two parts: the slice's
length s (straight from the region, the same top − bottom you already do for area), and the shape's
area formula in terms of s. Here's the procedure every problem follows:
1
Find the bounds. Set the curves equal — the intersection points are your limits of integration. (Same as area between curves.)
2
Find the slice length s. At a given x, it's the gap across the region: top − bottom (or right − left if slicing in y).
3
Write A(x). Plug s into the cross section's area formula — this is the step that decides everything. Get it wrong and the rest collapses.
4
Integrate A(x) over the bounds. That's the volume.
2 · The shape cheat sheet
All the area formulas in one place — keyed to the slice length s. Don't just memorize
them; hit "derive it" on any card to rebuild the formula from scratch, so you can always reconstruct it on
the exam. Every volume shown uses the same base region (y = 1 − x² over the x-axis) so you can see the one
thing that changes: the constant out front.
Every volume above is just a constant times the same integral, ∫(1−x²)² dx = 16/15 — because the base region and slice length s never change. Hit derive it on any card to see its full setup.
3 · A worked example, start to finish
The procedure on real curves. Watch the four steps — bounds, slice length, area, integrate — turn a region into a number.
Problem. The base of a solid is the region under y = √x from
x = 0 to x = 4. Cross sections perpendicular to the
x-axis are squares. Find the volume.
1
Bounds. The region runs from x = 0 to x = 4 — those are the limits.
2
Slice length. At each x, the region reaches from the x-axis up to the curve: s = top − bottom.
3
Area. Squares, so A = s² — and the √ disappears, which is why this one's clean.
4
Integrate. Antidifferentiate and evaluate.
Volume = 8 cubic units.
Swap to semicircles on the same base and only the constant changes:
— the cheat-sheet pattern in action.
When the slices are perpendicular to the y-axis
If the problem stacks slices along the y-axis instead, nothing about the
method changes — you just write everything in terms of y and integrate dy.
Problem. The base is bounded by the y-axis and
x = y² for 0 ≤ y ≤ 2. Cross sections perpendicular to the
y-axis are squares. Find the volume.
1
Pick the variable. Slices ⊥ to the y-axis stack along y — so integrate dy, with everything in terms of y.
2
Slice length. At height y, the region reaches from the y-axis (x = 0) out to x = y²: s = right − left.
3
Area. Squares again, so A = s².
4
Integrate over the y-bounds.
Volume = 32/5 cubic units.
Same four steps — only the variable switched from x to y. That's the whole dx-vs-dy decision.
4 · The traps that cost points
Almost every lost point on this topic comes from three specific mix-ups. Test yourself on each
before peeking.
① A semicircle's flat side spans the region — is that length the radius or the diameter?
Diameter. The slice spans all the way across, so the radius is half: r = s/2.
The tell: no /8 means you forgot to halve the diameter.
② "Isosceles right triangle" — is the region's slice the leg or the hypotenuse?
Depends — read it. The slice is either a leg or the hypotenuse:
The tell: same shape, different constant — check which side the region gives you.
③ The slices are perpendicular to the y-axis. Do you still integrate dx?
No — integrate dy. Slices ⊥ to the y-axis stack along y.
The tell: write s as right − left in terms of y. Same dx-vs-dy call as area between curves.
5 · Drill it — set up the volume
Six questions on picking the right formula, dodging the traps, and setting up the integral —
the parts that actually get tested. Full worked solution after each.
Score: 0 / 0
Next comes the other way to build a solid: spin a region around an axis instead of stacking shapes on it —
solids of revolution. Back to Unit 8.