2.4 - Surface areas of revolution

2.4 - Surface areas of revolution

Rotating a curve sweeps out a surface, and the distance from the axis changes the area swept out by each small piece. You will combine radius with arc length to find curved surface areas about either coordinate axis, using Cartesian or parametric equations.

The surface element

Why does a sloping line sweep out more area than a horizontal line with the same horizontal width and average distance from the axis? A surface strip depends on the actual sloping length, not just its horizontal projection.

Rotate a short straight piece about the horizontal axis. It makes a thin frustum, whose curved area is π(ρ1+ρ2)ℓ\pi(\rho_1+\rho_2)\ell: the mean of its two circumferences multiplied by its slant length. As the piece becomes shorter, both radii approach ρ\rho, and the slant length becomes the arc element ds\mathrm{d}s.

[DIAGRAM: asset_name: Radius and slant length determine the surface strip; asset_slug: cam9231-surface-strip; caption: A short curve element sweeps a frustum whose limiting area is circumference times arc length; description: A geometric side view shows the segment of y equals 1 plus x over two from x equals 1 to 1.6, its rotation about the horizontal x-axis, endpoint radii rho1 and rho2, and slant length ell. A separate clean formula panel shows frustum area pi times rho1 plus rho2 times ell approaching 2 pi rho ds. The end circles are outlines only; no end-disc area is included.]
A short curve element sweeps a frustum whose limiting area is circumference times arc length

Surface element of revolution

The curved area generated by a curve element is

dS=2πρ ds,\mathrm{d}S=2\pi\rho\,\mathrm{d}s,

where ρ≥0\rho\ge0 is its perpendicular distance from the axis of rotation. Integrate over a curve that sweeps the required surface once.

Not in the formula booklet: learn it.

This counts the swept curved surface. Flat closing discs are added only when the question separately asks for them as part of a total surface area.

Worked example 1: constant radius

Question. The line segment y=2y=2, 0≤x≤30\le x\le3, is rotated about the xx-axis. Find the area of the surface it generates.

Plan. Use radius 22 and ds=dx\mathrm{d}s=\mathrm{d}x.

S=∫032π(2) dx=12π.S=\int_0^3 2\pi(2)\,\mathrm{d}x=12\pi.

Every strip has the same circumference; no sloping correction is needed.

Answer. The generated curved surface area is 12π12\pi.

Check. Unrolling the cylinder gives a rectangle of width 2π(2)=4π2\pi(2)=4\pi and length 33, hence area 12π12\pi. Its two end discs have not been swept by the line segment.

Change from constant radius to a steadily increasing radius. Both the circumference and the sloping element now enter.

Worked example 2: a sloping generator

Question. The segment y=xy=x, 0≤x≤30\le x\le3, is rotated about the xx-axis. Find its generated surface area.

Plan. Use radius xx, and calculate the length element from y′=1y'=1.

ds=1+12 dx=2 dxS=2π∫03x2 dx=9π2\begin{aligned} \mathrm{d}s&=\sqrt{1+1^2}\,\mathrm{d}x=\sqrt2\,\mathrm{d}x \\ S&=2\pi\int_0^3x\sqrt2\,\mathrm{d}x=9\pi\sqrt2 \end{aligned}

Answer. S=9π2S=9\pi\sqrt2.

Check. This is a cone with base radius 33 and slant length 32+32=32\sqrt{3^2+3^2}=3\sqrt2. The geometric curved-area formula πrℓ\pi r\ell gives 9π29\pi\sqrt2 independently.

The opening question is answered by the length factor: replacing the sloping element by its horizontal projection would make each strip too small.

Watch out

2πρ dx2\pi\rho\,\mathrm{d}x uses horizontal width rather than arc length. In general use 2πρ1+(y′)2 dx2\pi\rho\sqrt{1+(y')^2}\,\mathrm{d}x.

Use a steeper generator to form the surface integral.

Revolving about either axis

The radius is always a distance from the chosen axis. For rotation about the xx-axis it is ∣y∣|y|; for rotation about the yy-axis it is ∣x∣|x|.

Cartesian surface areas

For a curve y=f(x)y=f(x), with a≤x≤ba\le x\le b, its swept curved area is

Sx=2π∫ab∣y∣1+(y′)2 dxSy=2π∫ab∣x∣1+(y′)2 dx\begin{aligned} S_x&=2\pi\int_a^b|y|\sqrt{1+(y')^2}\,\mathrm{d}x \\ S_y&=2\pi\int_a^b|x|\sqrt{1+(y')^2}\,\mathrm{d}x \end{aligned}

provided the chosen curve sweeps the required surface once. If x=g(y)x=g(y), use 1+(dx/dy)2 dy\sqrt{1+(\mathrm{d}x/\mathrm{d}y)^2}\,\mathrm{d}y instead, keeping the same radius rule. Split at sign changes to evaluate absolute values.

Not in the formula booklet: learn it.

The integration variable does not determine the axis. You can integrate with respect to xx when rotating about the yy-axis. Choose the variable that makes the curve and its limits easiest.

Worked example 3: a curve below the axis

Question. Find the area generated by rotating y=−cosh⁡xy=-\cosh x, 0≤x≤ln⁡20\le x\le\ln2, about the xx-axis.

Plan. Use the positive radius cosh⁡x\cosh x, then simplify the length element with the hyperbolic identity.

Since y′=−sinh⁡xy'=-\sinh x, 1+(y′)2=cosh⁡x>0\sqrt{1+(y')^2}=\cosh x>0. Thus

Sx=2π∫0ln⁡2cosh⁡2x dx=π[x+sinh⁡xcosh⁡x]0ln⁡2=π(ln⁡2+1516)\begin{aligned} S_x&=2\pi\int_0^{\ln2}\cosh^2x\,\mathrm{d}x \\ &=\pi[x+\sinh x\cosh x]_0^{\ln2} \\ &=\pi\left(\ln2+\frac{15}{16}\right) \end{aligned}

The endpoint values are sinh⁡(ln⁡2)=3/4\sinh(\ln2)=3/4 and cosh⁡(ln⁡2)=5/4\cosh(\ln2)=5/4.

Answer. Sx=π(ln⁡2+15/16)S_x=\pi(\ln2+15/16).

Check. Reflecting the curve in the xx-axis produces y=cosh⁡xy=\cosh x and exactly the same swept surface. A negative coordinate therefore cannot make the area negative. Also the arc length is ∫0ln⁡2cosh⁡x dx=3/4\int_0^{\ln2}\cosh x\,\mathrm{d}x=3/4, and 1≤ρ≤5/41\le\rho\le5/4; hence 3π/2≤Sx≤15π/83\pi/2\le S_x\le15\pi/8, which contains the result.

Change only the axis. The length element remains the same, but the radius is now xx.

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