Volumes with Cross Sections: Triangles and Semicircles: Study Guide (Calculus AB 8.8)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

Integrals are written in a compact form: ∫ (a to b) A(x) dx means the definite integral of A(x) from x = a to x = b. s always means the length of the side of the cross section that lies in the base: top − bottom for slices perpendicular to the x-axis, right − left (in y) for slices perpendicular to the y-axis.

Same integral, new shapes

Topic 8.7 built a solid on a base region R and sliced it into thin slabs. A slab at x has volume about A(x) Δx, and adding them gives

V = ∫ (a to b) A(x) dx

Nothing about that argument depended on the slice being a square. If the slices are triangles or semicircles, the method is the same. You only need the area of one slice in terms of s.

The area formulas, derived

Each shape below stands on a segment of length s in the base.

Equilateral triangle, side s. The height splits the base into two halves of s/2. By Pythagoras, height = √(s² − (s/2)²) = √(3s²/4) = (√3/2)s. So

A = (1/2) × s × (√3/2)s = (√3/4)s²

Isosceles right triangle with a leg in the base. Both legs have length s and meet at the right angle, so the other leg stands straight up. A = (1/2) × s × s = s²/2.

Isosceles right triangle with the hypotenuse in the base. The hypotenuse is s, so each leg is s/√2. A = (1/2)(s/√2)² = s²/4. Another way: the height to the hypotenuse is s/2, and (1/2) × s × (s/2) = s²/4.

Semicircle with its diameter in the base. The radius is s/2. A = (1/2)π(s/2)² = (π/8)s². Note s/2, not s: the side in the base is the diameter.

Other shapes. The same idea works for any shape you can find the area of.

• A triangle with base s and a stated height h: A = (1/2)sh.
• A full circle whose diameter spans the base: A = π(s/2)² = (π/4)s².
• A slice whose area is simply given as a function A(x): integrate it directly.

Cross section on side s	Area A	Constant c in A = cs²
Square (Topic 8.7)	s²	1
Equilateral triangle	(√3/4)s²	√3/4 ≈ 0.433
Isosceles right triangle, leg in base	s²/2	1/2
Isosceles right triangle, hypotenuse in base	s²/4	1/4
Semicircle, diameter in base	(π/8)s²	π/8 ≈ 0.393
Circle, diameter in base	(π/4)s²	π/4

Because A = c s² every time, V = c ∫ s² dx. Find ∫ s² dx once and you can get the volume for any of these shapes by multiplying by the right constant.

The method, step by step

1. Sketch the base and mark the direction of the slices.
2. Choose the variable. Perpendicular to the x-axis: x and dx. Perpendicular to the y-axis: y and dy.
3. Find the limits from the edges of the base.
4. Find s: top − bottom, or right − left written in y. Test a point to get the order right.
5. Read the shape carefully. Which part of the shape lies in the base: a side, a leg, the hypotenuse or the diameter? Write A in terms of s, and show where the constant comes from.
6. Write the integral in full, take the constant outside, evaluate, and give cubic units.

Worked example 1: equilateral triangles across x (no calculator)

Question. The base of a solid is the region enclosed by y = x² and y = 4. Cross sections perpendicular to the x-axis are equilateral triangles. Find the volume.

1. Limits. x² = 4 gives x = −2 or x = 2.
2. Side. The line y = 4 is on top (at x = 0, 4 > 0). s(x) = 4 − x².
3. Area. A(x) = (√3/4)(4 − x²)².
4. Integral. V = (√3/4) ∫ (−2 to 2) (4 − x²)² dx = (√3/4) ∫ (−2 to 2) (16 − 8x² + x⁴) dx
5. Evaluate. The antiderivative is 16x − 8x³/3 + x⁵/5. The integrand is even, so the integral is 2 × [value at x = 2] = 2 × (32 − 64/3 + 32/5) = 2 × 256/15 = 512/15.
6. Multiply by √3/4: V = (√3/4)(512/15) = 128√3/15.

Answer. V = 128√3/15 cubic units (about 14.780).

Check. If the slices were squares, the volume would be 512/15 ≈ 34.1. Equilateral triangles cover about 0.433 of a square on the same side, and 0.433 × 34.1 ≈ 14.8. That matches.

Worked example 2: semicircles across y (no calculator)

Question. The base of a solid is the region bounded by y = √x, the y-axis and the line y = 2. Cross sections perpendicular to the y-axis are semicircles with their diameters in the base. Find the volume.

1. Direction. Slices perpendicular to the y-axis: work in y with dy.
2. Limits. The region runs from y = 0 up to y = 2.
3. Side. At height y, a horizontal segment starts on the y-axis, x = 0, and ends on the curve. Rewrite y = √x as x = y². So s(y) = y² − 0 = y².
4. Area. The diameter is y², so the radius is y²/2. A(y) = (π/8)(y²)² = (π/8)y⁴.
5. Integral and value. V = (π/8) ∫ (0 to 2) y⁴ dy = (π/8)[y⁵/5] from 0 to 2 = (π/8)(32/5) = 4π/5

Answer. V = 4π/5 cubic units (about 2.513).

Typical error. Using s as the radius gives (π/2)(32/5) = 16π/5, four times too big. The side in the base is the diameter.

Worked example 3: a calculator-active volume

Question. R is the region enclosed by y = cos x and y = x². R is the base of a solid whose cross sections perpendicular to the x-axis are isosceles right triangles with the hypotenuse in R. Find the volume. (Calculator allowed.)

1. Limits. cos x = x² needs a calculator. By symmetry the solutions are x = −C and x = C, with C ≈ 0.824. Store C.
2. Top curve. At x = 0, cos 0 = 1 and 0² = 0, so cos x is on top. s(x) = cos x − x².
3. Area. Hypotenuse in the base: A(x) = s²/4 = (cos x − x²)²/4.
4. Integral. Write it first: V = ∫ (−C to C) (cos x − x²)²/4 dx
5. Evaluate with stored C: V ≈ 0.219 cubic units.

Sense check. The widest slice is at x = 0, with s = 1 and A = 1/4. The base is about 1.648 wide, so the volume must be less than (1/4)(1.648) ≈ 0.412. It is. With the leg in the base instead, the answer would double, to about 0.437.

Other shapes and given areas

Circles across the base. Suppose the base is the triangle under y = x for 0 ≤ x ≤ 2, and each slice perpendicular to the x-axis is a full circle whose diameter spans the base (like a horn or funnel). Then A(x) = (π/4)x², and V = (π/4) ∫ (0 to 2) x² dx = (π/4)(8/3) = 2π/3.

An area function. Sometimes you are told the cross-sectional area directly. If a solid lies between x = 0 and x = 3 and the slice at x has area A(x) = 12/(x + 1)², then V = ∫ (0 to 3) 12/(x + 1)² dx = 12[−1/(x + 1)] from 0 to 3 = 12(1 − 1/4) = 9 cubic units. No base region is needed: the area function already carries all the information.

Common misconceptions

• Using s as the radius of a semicircle. The side in the base is the diameter, so r = s/2 and A = (π/8)s², not (π/2)s².
• Mixing up the two right-triangle cases. Leg in the base: s²/2. Hypotenuse in the base: s²/4.
• Wrong equilateral constant. It is √3/4, not √3/2 (that is the height factor).
• Squaring each boundary separately. Find s = top − bottom first, then square: (f − g)², not f² − g².
• Forgetting to square s. Every area here is a constant times s².
• Mixing variables. Slices perpendicular to the y-axis need s in terms of y, y-limits and dy.
• Confusing this with the disc method. Here the semicircles stand on the base. In Topic 8.9 the circles come from spinning a region about an axis, and the radius is measured from that axis.

Where this leads

Topic 8.9 uses the same V = ∫ A dx with circular slices of area πr², where the radius comes from revolving a region around an axis. Continue with Volume with Disc Method: Revolving Around the x- or y-Axis, or look back at Volumes with Cross Sections: Squares and Rectangles. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.