E7 - Surface area of revolution | AQA A-Level Further Mathematics

Rotating a curve produces a surface, whose thin bands have area equal to circumference times slant length in the limit. You will derive this rule, use Cartesian and parametric curves about either coordinate axis, and select nonnegative radii and branches that count the surface once.

Derive the frustum-element formula

Why does rotating a semicircle of radius 2 give a sphere with area $16\pi$ , even though the semicircle's length is only $2\pi$ ? Different points travel around circles of different radii. Surface area depends on both the generating curve's length and each point's distance from the axis.

Approximate a short curve segment by a straight chord. Rotating that chord about an axis produces a narrow cone frustum. If its endpoint radii are $r_1,r_2$ and its slant length is $\ell$ , its curved area is $\pi(r_1+r_2)\ell$ : average circumference times slant length.

To derive this, suppose $r_2>r_1\ge0$ and extend the straight sides to form two similar cones with slant heights $L_2,L_1$ . Similarity and subtraction give $L_2/r_2=L_1/r_1$ when $r_1>0$ , and $L_2-L_1=\ell$ . Thus $L_2=r_2\ell/(r_2-r_1)$ and $L_1=r_1\ell/(r_2-r_1)$ , also valid when the smaller cone has radius and height zero. Subtract their curved areas:

\pi r_2L_2-\pi r_1L_1 =\frac{\pi(r_2^2-r_1^2)\ell}{r_2-r_1} =\pi(r_1+r_2)\ell

Equal radii give the cylindrical band area $2\pi r\ell$ directly.

As the segment shrinks, both endpoint radii approach $r$ , while the chord approaches arc element $\mathrm{d}s$ . Thus $\mathrm{d}S=2\pi r\,\mathrm{d}s$ .

[DIAGRAM: asset_name: A narrow surface band; asset_slug: aqa7367-surface-band; caption: A frustum band approaches circumference times slant length; description: An axial cross-section shows two nearby points on a rising generating curve at radii r1 and r2 from the horizontal rotation axis. Their chord has slant length ell. A right schematic shows the frustum formed by rotating the chord, with endpoint circular rims and the curved band shaded. Its area pi times r1 plus r2 times ell is shown, tending to 2 pi r ds for a short curve element.]
A frustum band approaches circumference times slant length

Surface area from a generating curve

For a full rotation about the $x$ -axis, the radius is $r=|y|$ , so

S=2\pi\int r\,\mathrm{d}s =2\pi\int_a^b |y|\sqrt{1+(y')^2}\,\mathrm{d}x

Use increasing $x$ -bounds and a curve branch that generates each surface patch once. The booklet supplies the Cartesian $x$ -axis formula with $y$ as radius, for a branch with $y\ge0$ .

This gives the curved surface swept out by the curve. Circular end caps are separate areas unless included in the question's surface.

Worked example 1: a cone's curved surface

Question. The segment $y=x$ , $0\le x\le1$ , rotates about the $x$ -axis. Find the generated surface area.

Plan. Use radius $x$ and length factor $\sqrt{1+1^2}=\sqrt2$ .

S=2\pi\int_0^1 x\sqrt2\,\mathrm{d}x =2\pi\sqrt2\left[\frac{x^2}{2}\right]_0^1 =\pi\sqrt2

Answer. $\pi\sqrt2$ square units, excluding the base disc.

Check. The cone has radius 1 and slant height $\sqrt2$ , so its curved area $\pi r\ell$ is $\pi\sqrt2$ .

Keep the horizontal interval and double the gradient. Both the radius and the slant-length factor change.

Worked example 2: a steeper cone

Question. The segment $y=2x$ , $0\le x\le1$ , rotates about the $x$ -axis. Find the generated area.

Plan. Use radius $2x$ and length factor $\sqrt{1+2^2}=\sqrt5$ .

S=2\pi\int_0^1 2x\sqrt5\,\mathrm{d}x =2\pi\sqrt5

Answer. $2\pi\sqrt5$ square units.

Check. The cone has radius 2 and slant height $\sqrt{1^2+2^2}=\sqrt5$ . Its curved area is $\pi(2)\sqrt5$ , agreeing independently.

A length factor distinguishes surface area from both volume and ordinary area under a graph.

Watch out

Wrong: $S=2\pi\int y\,\mathrm{d}x$ . This omits the slant length. Use $2\pi\int r\sqrt{1+(y')^2}\,\mathrm{d}x$ ; add end caps only if asked for them.

Use a new cone slope for your own calculation.

Cartesian and parametric curves about either axis

Surface area always uses the same two ingredients: distance from the chosen axis and distance along the curve. Changing the axis changes the radius; changing the parameter changes the speed.

Coordinate forms of surface area

For $\alpha<\beta$ , write $v(t)=\sqrt{x'(t)^2+y'(t)^2}$ . Then

S_x=2\pi\int_\alpha^\beta |y(t)|v(t)\,\mathrm{d}t, \qquad S_y=2\pi\int_\alpha^\beta |x(t)|v(t)\,\mathrm{d}t

For a Cartesian curve $y=f(x)$ , the $y$ -axis form is $S_y=2\pi\int_a^b|x|\sqrt{1+(y')^2}\,\mathrm{d}x$ . If instead $x=g(y)$ , it is $2\pi\int |x|\sqrt{1+(\mathrm{d}x/\mathrm{d}y)^2}\,\mathrm{d}y$ .

The booklet prints Cartesian and parametric forms about the $x$ -axis with nonnegative $y$ as radius. The $y$ -axis forms follow by changing the radius; they are not separately printed.

Use a branch and interval that generate the required surface once. In a parametric calculation, speed is nonnegative regardless of direction; there is no requirement that $x$ increase. A repeated or opposite branch can still sweep out surface already counted.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Core Pure

Mechanics

Statistics

Discrete Mathematics

E7 - Surface area of revolution

Derive the frustum-element formula

Cartesian and parametric curves about either axis

Continue this lesson.

Learn faster with Nova