8.06b - Arc lengths and surface areas

Arc length from a cartesian equation

A sphere of radius $2$ fits exactly inside a cylinder of radius $2$ and height $4$. Which has the larger area: the sphere's surface or the cylinder's curved surface?

Volume formulae will not settle that question: a curved surface is made from sloping lengths, rather than circular discs. Begin with the length of the curve that will generate it. A straight segment with horizontal change $3$ and vertical change $4$ has length $5$, not $3$ or $7$. The same Pythagoras idea applies to very small pieces of a smooth curve.

[DIAGRAM: asset_name: A curve is approximated by short sloping segments; asset_slug: h245_8_06_arc_element; caption: Arc length adds sloping lengths, with both horizontal and vertical changes contributing; description: Left panel shows a smooth rising curve approximated by five chords. Right panel enlarges one short chord and its right-angled triangle, labelled delta x, delta y and delta s. Beneath it, the limiting expression is ds equals square root of one plus derivative squared times dx.]

For a small horizontal change $\Delta x>0$, the chord has length $\sqrt{(\Delta x)^2+(\Delta y)^2}$. Factoring out $\Delta x$, adding the chords and taking a smooth-curve limit replaces their slope by $dy/dx$. The local length element is $ds=\sqrt{1+(dy/dx)^2}\,dx$. This also explains the square on the derivative: a downward slope contributes just as much length as its upward reflection.

If the equation is naturally $x=g(y)$, interchange the roles of the coordinates: $\displaystyle s=\int_c^d\sqrt{1+(dx/dy)^2}\,dy$, for $c<d$. This equivalent form follows from the same Pythagoras argument; the booklet prints the $y=f(x)$ version. Use the form with the simpler derivative, and split into pieces if one equation cannot describe the entire arc.

Change only the roles of $x$ and $y$. Reflecting in $y=x$ preserves length, but changes the convenient integration variable.

Arc length is not the signed change in either coordinate.

Keep the same reflected curve and change its endpoint.

Arc length from parametric equations

A parameter can describe an arc even when its horizontal or vertical direction reverses, so calculate the small length from both parametric derivatives together.

If $x=x(t)$ and $y=y(t)$, a small positive parameter change gives coordinate changes approximately $x'(t)\Delta t$ and $y'(t)\Delta t$. Their sloping length is approximately $\sqrt{x'(t)^2+y'(t)^2}\Delta t$. Neither derivative has to be positive.

The quantity under the integral is the rate at which length is traced. Factor it before taking the square root: an apparent quartic may be a perfect square.

Change only the lower parameter limit to $-1$. The vertical coordinate now decreases and then increases, but both halves of the arc still contribute positive length.

A reversal of a coordinate is harmless; retracing the same arc changes what the integral counts. For example, $x=t^2$, $y=0$, $-1\le t\le1$, traverses the segment from $x=1$ to $0$ and back. Its tracing integral is $\int_{-1}^1 2|t|\,dt=2$, but the segment itself has length $1$. Choose an interval that traces the required arc once.

The same polynomial curve remains straightforward beyond $t=1$, where its horizontal coordinate starts increasing again.

Cartesian surfaces about the x-axis

Rotating a short sloping segment produces a thin conical band, so its curved area depends on the sloping length rather than the horizontal width.

A conical frustum has curved area $\pi(r_1+r_2)l$, where $l$ is its sloping length. For a very thin band the two radii approach the same radius $r$, giving $dS=2\pi r\,ds$. This local circumference-times-length relation, derived here and not separately printed in the booklet, explains every surface formula in the lesson.

[DIAGRAM: asset_name: Sloping bands and their distances from the axis; asset_slug: h245_8_06_surface_bands; caption: Each curved band contributes circumference times sloping length, and its radius is perpendicular to the rotation axis; description: Left panel shows y equals x plus one from x zero to two, a highlighted short segment, its vertical radius y to the x-axis, and a schematic conical band labelled 2 pi y ds. Right panel shows the reflected branches x equals plus or minus square root of 2y, with horizontal distances to the y-axis both labelled absolute x.]

The frustum formula is assumed geometry and is not printed in the booklet. In the limit, the surface of a smooth curve is found by integrating these bands. For rotation about the $x$-axis, the radius is $|y|$.

A straight generating line provides a useful check because its surface is a cone or a frustum.

Change only the vertical position of the line, from $y=x$ to $y=x+1$. The sloping length stays the same, but every circumference is larger.

If the question asks for the total boundary area of the closed frustum, add its two flat discs: here their total area is $\pi(1^2)+\pi(3^2)=10\pi$. The phrase “surface generated by the curve” asks for the curved area alone.

Shift the line one unit farther from the same axis.

Cartesian surfaces about the y-axis

For rotation about the $y$-axis, the radius is the horizontal distance $|x|$, and the booklet's cartesian form uses $y$ as the integration variable.

The same $dS=2\pi r\,ds$ argument applies. Expressing the curve as $x=g(y)$ makes its sloping-length element $\sqrt{1+(dx/dy)^2}\,dy$. Both the radius and the derivative must be functions of $y$.

Alternatively, if $y=f(x)$ gives a simpler calculation, keep $ds=\sqrt{1+(dy/dx)^2}\,dx$ and integrate $2\pi|x|\,ds$ using $x$-limits. That equivalent choice follows from the same length element; it is not a separate printed cartesian formula. The axis chooses the radius, while your representation chooses the integration variable.