MathematicsGrade 11–13

Solids of revolution

The area under a graph rotates about the x-axis and becomes a solid. Its volume is calculated with thin discs.

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What is it about?

If the region between the graph of ff and the xx-axis on [a,b][a, b] turns once about the xx-axis, a solid of revolution appears: a rectangle gives a cylinder, a triangle a cone, a semicircle a sphere.

The cross-section at xx is a circle with radius ∣f(x)∣|f(x)| and area A(x)=π(f(x))2A(x) = \pi \big(f(x)\big)^2. Cutting [a,b][a, b] into nn strips of width Δx\Delta x gives nn cylindrical discs with total volume Vn=∑πri2 ΔxV_n = \sum \pi r_i^2 \, \Delta x. With inner discs the sum is too small, with outer discs too large; for n→∞n \to \infty both tend to

V=π∫ab(f(x))2 dx.V = \pi \int_a^b \big(f(x)\big)^2 \, dx .

Checks: f(x)=r2−x2f(x) = \sqrt{r^2 - x^2} on [−r,r][-r, r] gives 43πr3\tfrac{4}{3}\pi r^3, f(x)=rhxf(x) = \tfrac{r}{h}x on [0,h][0, h] gives 13πr2h\tfrac{1}{3}\pi r^2 h, f(x)=xf(x) = \sqrt{x} on [0,4][0, 4] gives 8π8\pi.

Try it

  1. Press “Rotate” and watch the green region sweep out the solid. Quarter turn
  2. Move the orange cross-section. Where is A(x)A(x) largest? Sphere
  3. Cut the cone into 4 inner discs, then outer discs, and double nn. Cone, 4 discs
  4. Gabriel’s horn: move bb further and further to the right. Horn

Tasks

Task 1: Cone

f(x)=0.5xf(x) = 0.5x turns about the xx-axis on [0,4][0, 4]. Find the volume and check it with the cone formula. Load task

Show solution

V=π∫040.25x2 dx=163π≈16.76V = \pi \int_0^4 0.25x^2 \, dx = \tfrac{16}{3}\pi \approx 16.76; cone: 13π⋅22⋅4=163π\tfrac{1}{3}\pi \cdot 2^2 \cdot 4 = \tfrac{16}{3}\pi.

Task 2: Sphere

Use f(x)=9−x2f(x) = \sqrt{9 - x^2} to find the volume of a sphere with r=3r = 3. Load task

Show solution

V=π∫−33(9−x2) dx=36π≈113.10=43π⋅33V = \pi \int_{-3}^{3} (9 - x^2) \, dx = 36\pi \approx 113.10 = \tfrac{4}{3}\pi \cdot 3^3.

Task 3: Gabriel’s horn

Find the volume for f(x)=1xf(x) = \tfrac{1}{x} on [1,b][1, b] and let b→∞b \to \infty. Load task

Show solution

V(b)=π(1−1b)V(b) = \pi\left(1 - \tfrac{1}{b}\right), so V(10)=0.9π≈2.83V(10) = 0.9\pi \approx 2.83 and V→πV \to \pi: an infinitely long solid with finite volume.

Ideas for the classroom

  • Year 13, applications of the integral; links to the volume formulas from year 10. Disc radius: midpoint value, smallest or largest value in each strip.
  • Typical mistakes: forgetting to square ff (doubling ff multiplies the volume by 4), squaring the integral instead of the integrand, expecting negative volumes for graphs below the axis.