Mathematics.

applications of integration

Surface Area of Revolution

Calculus II45 minDifficulty5 out of 10

You should know: arc length, integral

Overview

When a curve is revolved around an axis, it sweeps out a surface. The surface area of revolution formula integrates the circumference of infinitesimal rings along the curve. This leads to elegant results such as recovering the sphere's surface area S = 4*pi*r^2 and the surprising Gabriel's Horn paradox, where infinite surface area coexists with finite volume.

Intuition

Imagine revolving a tiny arc element ds of length ds around the x-axis. It sweeps out a thin ring (frustum) with circumference 2*pi*f(x). The surface area is the sum (integral) of all these ring circumferences times the arc length element: S = integral 2*pi*f(x) ds. Since ds = sqrt(1 + (f')^2) dx, we get the standard formula.

Formal Definition

Definition

If f is a smooth non-negative function on [a,b], the surface area generated by revolving the curve y = f(x) around the x-axis is given by the integral below. For rotation around the y-axis, replace f(x) with x.

S=2π∫abf(x)1+[f′(x)]2 dxS = 2\pi \int_a^b f(x)\sqrt{1 + [f'(x)]^2}\, dx
Surface area around the x-axis
S=2π∫abx1+[f′(x)]2 dxS = 2\pi \int_a^b x\sqrt{1 + [f'(x)]^2}\, dx
Surface area around the y-axis
S=2π∫αβy(t)[x′(t)]2+[y′(t)]2 dtS = 2\pi \int_\alpha^\beta y(t)\sqrt{[x'(t)]^2 + [y'(t)]^2}\, dt
Parametric form (rotation around x-axis)

Notation

NotationMeaning
SSSurface area of the solid of revolution
ds=1+(f′)2 dxds = \sqrt{1+(f')^2}\,dxArc length element along the curve
f(x)f(x)Distance from the axis (radius of ring)

Theorems

Theorem 1: Surface Area Formula (x-axis rotation)
Fory=f(x)>=0smoothon[a,b],thesurfaceareaofrevolutionaroundthex−axisisS=2∗pi∗integralabf(x)∗sqrt(1+[f′(x)]2)dx.For y = f(x) >= 0 smooth on [a,b], the surface area of revolution around the x-axis is S = 2*pi * integral_a^b f(x)*sqrt(1 + [f'(x)]^2) dx.
Theorem 2: Sphere via Revolution
Rotatingthesemicircley=sqrt(r2−x2)on[−r,r]aroundthex−axisgivesasphereofradiusrwithsurfaceareaS=4∗pi∗r2,recoveringArchimedes′formula.Rotating the semicircle y = sqrt(r^2 - x^2) on [-r,r] around the x-axis gives a sphere of radius r with surface area S = 4*pi*r^2, recovering Archimedes' formula.
Theorem 3: Gabriel's Horn Paradox
Thesurfaceformedbyrotatingy=1/xforx>=1aroundthex−axishasinfinitesurfacearea(integral1/xdiverges)butfinitevolumepi(integral1/x2converges).ThisistheGabriel′sHornparadox:youcanfillitwithpaintbutcannotpaintitsinside.The surface formed by rotating y = 1/x for x >= 1 around the x-axis has infinite surface area (integral 1/x diverges) but finite volume pi (integral 1/x^2 converges). This is the Gabriel's Horn paradox: you can fill it with paint but cannot paint its inside.

Worked Examples

  1. 1

    Compute f'(x) = -x/sqrt(r^2-x^2).

    f′(x)=−xr2−x2f'(x) = \frac{-x}{\sqrt{r^2-x^2}}
  2. 2

    Compute the integrand factor:

    1+[f′(x)]2=1+x2r2−x2=r2r2−x2=rr2−x2\sqrt{1+[f'(x)]^2} = \sqrt{1 + \frac{x^2}{r^2-x^2}} = \sqrt{\frac{r^2}{r^2-x^2}} = \frac{r}{\sqrt{r^2-x^2}}
  3. 3

    So f(x)*sqrt(1+(f')^2) = sqrt(r^2-x^2) * r/sqrt(r^2-x^2) = r.

    f(x)1+(f′)2=rf(x)\sqrt{1+(f')^2} = r
  4. 4

    Integrate from -r to r:

    S=2π∫−rrr dx=2πr⋅2r=4πr2S = 2\pi \int_{-r}^{r} r\, dx = 2\pi r \cdot 2r = 4\pi r^2

✓ Answer

S = 4*pi*r^2, confirming Archimedes' formula for the surface area of a sphere.

Practice Problems

Easyapplication

Find the surface area generated by rotating y = sqrt(x) for x in [0,4] around the x-axis.

Mediumapplication

Verify that Gabriel's Horn (y = 1/x, x >= 1, rotated around x-axis) has infinite surface area.

Easyapplication

Find the surface area of the cone formed by rotating y = (r/h)*x for x in [0,h] around the x-axis.

Common Mistakes

Common Mistake

The surface area formula is the same as the volume formula (using disks).

Volume uses pi*[f(x)]^2 dx (disk area); surface area uses 2*pi*f(x)*sqrt(1+(f')^2) dx (ring circumference times arc length). The arc length factor sqrt(1+(f')^2) is essential.

Common Mistake

Forgetting the arc length factor: writing S = 2*pi*integral f(x) dx.

The rings are slanted, not vertical. The correct element is ds = sqrt(1+(f')^2) dx, not just dx.

Quiz

The surface area formula S = 2*pi*integral f(x)*sqrt(1+(f')^2) dx comes from summing:
Rotating y = sqrt(r^2 - x^2) for x in [-r,r] around the x-axis gives surface area:
Gabriel's Horn is paradoxical because:
For rotation around the y-axis (not x-axis), the surface area formula uses:

Historical Background

Archimedes computed the surface area of a sphere by inscribing and circumscribing solids in the 3rd century BCE, obtaining S = 4*pi*r^2 -- one of his proudest results. The integral formula for surface area of revolution awaited the development of calculus by Newton and Leibniz in the 17th century. Gabriel's Horn (Torricelli's trumpet) was described by Evangelista Torricelli in 1643 and immediately provoked philosophical debate about infinity.

  1. 225 BCE

    Archimedes derives the surface area of a sphere geometrically

    Archimedes

  2. 1643

    Torricelli describes Gabriel's Horn (1/x rotated around x-axis for x >= 1)

    Evangelista Torricelli

  3. 1690s

    Newton and Leibniz develop calculus, enabling integral formulas for surface area

    Isaac Newton, Gottfried Wilhelm Leibniz

Summary

  • The surface area of revolution formula is S = 2*pi * integral_a^b f(x)*sqrt(1+[f'(x)]^2) dx for rotation around the x-axis.
  • Rotating the semicircle y = sqrt(r^2-x^2) recovers the sphere surface area S = 4*pi*r^2.
  • Gabriel's Horn (y=1/x, x>=1) has infinite surface area but finite volume pi -- the Gabriel's Horn paradox.
  • The arc length element ds = sqrt(1+(f')^2) dx is the key factor distinguishing surface area from simpler integrals.

References

  1. BookStewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage Learning.
  2. BookThomas, G. B. et al. (2014). Thomas' Calculus (13th ed.). Pearson.
  3. WebsiteWeisstein, E. W. Surface of Revolution. MathWorld.