How to Find Volumes with Cross Sections Triangles and Semicircles

To find volumes with cross sections triangles and semicircles, set up a definite integral of the form V = ∫ A(x) dx over the base region, where A(x) is the area of a slice perpendicular to your chosen axis: use A = (√3/4)s² for equilateral triangles and A = (π/8)s² for semicircles built on a chord of length s, then match s to the base width at each x and integrate from the left edge to the right edge.

Core Volume Formula

The slicing method treats a solid as a stack of thin slabs. If a slice perpendicular to the x-axis at position x has area A(x), then the slab has volume A(x)·Δx, and summing infinitely many slabs gives

V = ∫ₐᵇ A(x) dx

Here A(x) is the cross-sectional area function, and [a, b] is the interval over which the solid extends along the axis. The classic use case is a solid whose base is a region in the xy-plane — such as the region between a curve and the x-axis — with cross sections cut perpendicular to that plane.

Two setup decisions drive everything:

• Axis of slicing. Slicing perpendicular to the x-axis means every area formula is written in terms of x. Slicing perpendicular to the y-axis means everything is in terms of y, with limits read off the y-extent.
• The chord length s(x). Each cross section sits on a chord of the base region. For a base bounded above by y = f(x) and below by y = g(x), that chord has length s(x) = f(x) − g(x). For a base bounded by a single curve above the x-axis, s(x) = f(x).

The shape of the cross section only changes the area formula, not the integral structure. That is why triangles and semicircles are usually taught together — the setup is identical and only A(x) differs. Khan Academy’s treatment of solids with known cross sections follows exactly this pattern.

Students often lose points by hunting for the “right” formula. The formula is given or standard; the real work is expressing the chord length correctly in one variable. Once s(x) is correct, substituting into the area formula and integrating is mechanical.

Triangle Cross Sections

An equilateral triangle with side length s has area

A = (√3/4)s²

This comes from the standard triangle area ½bh with height h = (√3/2)s. If the cross sections are equilateral triangles whose bases are chords of length s(x), then

A(x) = (√3/4)·[s(x)]²

and V = (√3/4)∫ₐᵇ [s(x)]² dx.

For isosceles right triangles, the area depends on which side is the chord:

• Chord as one leg: A = ½s²
• Chord as the hypotenuse: A = ½(s/√2)² = s²/4

For a general isosceles triangle with base s and altitude h, A = ½sh; if the altitude is given as a fixed multiple of s, substitute to keep everything in one variable.

Reading the base correctly

Suppose the base is the region bounded by y = √x, the x-axis, and x = 4. Then s(x) = √x, and the equilateral-triangle area is A(x) = (√3/4)(√x)² = (√3/4)x. The integral becomes

V = (√3/4)∫₀⁴ x dx = (√3/4)·8 = 2√3

or about 3.464 cubic units. Notice how the square root disappeared once squared — a common and welcome simplification.

When the base is bounded by two curves, say y = x and y = x² on [0, 1], the chord is s(x) = x − x². Every triangle formula then squares that difference, which produces a polynomial you can integrate term by term.

Semicircle Cross Sections

A circle of radius r has area πr², so a semicircle has area

A = ½πr²

If the diameter of the semicircle is the chord of length s, then r = s/2 and

A = ½π(s/2)² = (π/8)s²

This is the formula to memorize for semicircular cross sections built on a chord. With diameter s(x),

A(x) = (π/8)[s(x)]², so V = (π/8)∫ₐᵇ [s(x)]² dx

The factor π/8 is easy to misremember as π/4 (that would be a full circle) or π/2 (that would use s as a radius). Anchoring on “semicircle on a diameter of length s” keeps it straight. MIT OpenCourseWare’s calculus materials on applications of integration walk through this same diameter-to-area chain.

The limits of integration are always the extent of the base along the slicing axis — not the radius, not the height. If the base runs from x = 0 to x = 3, the limits are 0 and 3 regardless of the cross-section shape. If you switch to slicing perpendicular to the y-axis, invert the curve to x = f⁻¹(y) and read limits off the y-values.

One practical note: semicircular cross sections stacked along a curve can produce solids whose top surface is curved, but the volume integral does not care. You only need area as a function of position.

Worked Example

Semicircles. Base: the region between y = x and y = x² on [0, 1]. Cross sections perpendicular to the x-axis are semicircles with diameters in the base plane.

• Chord: s(x) = x − x²
• Area: A(x) = (π/8)(x − x²)²
• Volume: V = (π/8)∫₀¹ (x − x²)² dx = (π/8)∫₀¹ (x² − 2x³ + x⁴) dx

That integral is 1/3 − 1/2 + 1/5 = 1/30. So V = (π/8)(1/30) = π/240 ≈ 0.0131 cubic units.

For an equilateral-triangle example on the same base, the setup is identical except for the constant: V = (√3/4)∫₀¹ (x − x²)² dx = (√3/4)(1/30) = √3/120. Same chord, same integral, different shape factor.

Setup checklist

1. Sketch the base region and mark the slicing axis.
2. Write the chord length s as a function of the slicing variable.
3. Choose the area formula for the stated shape and substitute s.
4. Set limits from the base’s extent along that axis.
5. Integrate, then sanity-check that the units are cubic and the answer is positive.

If the problem says “cross sections perpendicular to the y-axis,” swap x and y everywhere in steps 2 through 4. The arithmetic changes; the logic does not.

Key Takeaways

• Volume by slicing is V = ∫ₐᵇ A(x) dx, where A(x) is the cross-sectional area at position x.
• Equilateral triangle cross sections use A = (√3/4)s²; semicircles on a diameter s use A = (π/8)s².
• The chord length s(x) is the base width at each x — for two bounding curves, s(x) = f(x) − g(x), and you square it in every area formula.
• Limits of integration come from the base region’s extent along the slicing axis, never from the cross-section dimensions.
• Choose the axis first, express everything in that one variable, and only then integrate.