Volumes with Known Cross Sections

A base and its slices

Start with a flat region on the floor: everything under $y=\sqrt{x}$y = √x from x = 0 to 4. Cut it into thin strips across the x-axis. The strip at x runs from the axis up to the curve, so it is $\sqrt{x}$√x long.

Now stand a square on every strip, with the strip as one side. The squares are small near the origin and grow to side 2 at x = 4, and together they make a solid. Nothing is spun: the slices are given.

The base, seen from above: the region under $y=\sqrt{x}$y = √x from 0 to 4. The strip at x is $\sqrt{x}$√x long, and a square of that side stands on it.

The slice at x, face on: a square of side $s=\sqrt{x}$s = √x, so its area is $s^2=x$s² = x.

Face area times thickness

Write A(x) for the area of the slice at x. Here $A\left(x\right)={\left(\sqrt{x}\right)}^2=x$A(x) = (√x)² = x. A slab of thickness dx has volume A(x) dx, its face area times its thickness, and adding the slabs gives $V=\int A\left(x\right)dx$V = ∫ A(x) dx.

For the squares, V is the integral of x from 0 to 4, which is $\frac{16}{2}=8$16/2 = 8 cubic units. No $\pi$π appears, because no face is a circle.

The disc method is this formula with circular faces: a disc of radius y has $A=\pi y^2$A = πy², and $V=\int \pi y^{2}dx$V = ∫ πy² dx.

Other slice shapes

Change the shape standing on each strip and only A(x) changes. With s the strip’s length:

A rectangle of height h has area sh. With h = 2 on the same base, $A\left(x\right)=2\sqrt{x}$A(x) = 2√x, and V is the integral of $2\sqrt{x}$2√x from 0 to 4, which is $2\times \frac{2}{3}\times 4\sqrt{4}=2\times \frac{2}{3}\times 8=\frac{32}{3}$2 × ⅔ × 4√4 = 2 × ⅔ × 8 = 32/3.

An equilateral triangle of side s stands $\left(\frac{\sqrt{3}}{2}\right)s$(√3/2)s tall, so its area is $\frac{1}{2}\times s\times \left(\frac{\sqrt{3}}{2}\right)s=\left(\frac{\sqrt{3}}{4}\right)s^2$½ × s × (√3/2)s = (√3/4)s². On the same base, $A\left(x\right)=\left(\frac{\sqrt{3}}{4}\right)x$A(x) = (√3/4)x and $V=\left(\frac{\sqrt{3}}{4}\right)\times 8=2\sqrt{3}$V = (√3/4) × 8 = 2√3, about 3.46.

A semicircle on diameter s has radius $\frac{s}{2}$s/2, so its area is half of $\pi {\left(\frac{s}{2}\right)}^2$π(s/2)², which is $\left(\frac{\pi }{8}\right)s^2$(π/8)s². On the same base, $A\left(x\right)=\left(\frac{\pi }{8}\right)x$A(x) = (π/8)x and $V=\left(\frac{\pi }{8}\right)\times 8=\pi$V = (π/8) × 8 = π.

A base between two curves

If the base lies between two curves, the strip at x runs from the lower curve to the upper, so its length is upper − lower. For the base between y = x and $y=x^2$y = x² on [0, 1], $s=x-x^2$s = x − x².

With squares, $A\left(x\right)={\left(x-x^2\right)}^2=x^2-2x^3+x^4$A(x) = (x − x²)² = x² − 2x³ + x⁴, and V is its integral from 0 to 1: $\frac{1}{3}-\frac{1}{2}+\frac{1}{5}=\frac{1}{30}$1/3 − 1/2 + 1/5 = 1/30.

A base between y = x and $y=x^2$y = x², from (0, 0) to (1, 1). The strip at x runs from the parabola up to the line, so its length is $x-x^2$x − x².

A pyramid, slice by slice

A pyramid with a square base of side b and height h can be cut into square slices parallel to the base. Measure x down from the apex. The side grows in proportion to x, so at depth x it is $\frac{bx}{h}$bx/h, and $A\left(x\right)=\frac{b^2{x}^2}{h^2}$A(x) = b²x²/h².

Then V is the integral of $\frac{b^2{x}^2}{h^2}$b²x²/h² from 0 to h, which is $\left(\frac{b^2}{h^2}\right)\times \frac{h^3}{3}=\frac{b^{2}h}{3}$(b²/h²) × h³/3 = b²h/3: a third of the box around it. With b = 6 and h = 4 that is $36\times \frac{4}{3}=48$36 × 4/3 = 48.

The usual mistakes

Putting in $\pi$π. $8\pi$8π for the squares on $\sqrt{x}$√x carries a factor $\pi$π that belongs only to circular faces.

Squaring the area. The slab is A(x) dx; $A{\left(x\right)}^2$A(x)² is not a volume.

Skipping the divide. The integral of x from 0 to 4 is 8, not 16.

Using the whole circle for a semicircle. On diameter s the circle has area $\left(\frac{\pi }{4}\right)s^2$(π/4)s², and the semicircle half of that, $\left(\frac{\pi }{8}\right)s^2$(π/8)s². Taking s as the radius gives $\left(\frac{\pi }{2}\right)s^2$(π/2)s², four times too much.

Giving an equilateral triangle height s. Its height is $\left(\frac{\sqrt{3}}{2}\right)s$(√3/2)s, so its area is $\left(\frac{\sqrt{3}}{4}\right)s^2$(√3/4)s², not $\frac{1}{2}{s}^2$½s².

Measuring the side from the axis when the base lies between two curves. The side is upper − lower.

A marquee

In the application below, a marquee stands over a round floor on triangular ribs. Each rib is a cross section whose base is the width of the floor at that point, so A(x) changes across the floor, and integrating it gives the air under the canvas.