Arc Length & Surface Area

Arc Length in 2-D

Consider the parabola $\mathbf{r}(t) = (t,\, t^2)$ for $t \in [0, 2]$. We approximate its length by $n$ straight segments, then take the limit.

Each segment has length $\sqrt{(\Delta x_i)^2 + (\Delta y_i)^2}$. In the limit:

For $\mathbf{r}(t)=(t, t^2)$: $\mathbf{r}'(t)=(1, 2t)$, so $|\mathbf{r}'(t)|=\sqrt{1+4t^2}$.

Arc Length in 3-D

The same idea extends to space curves. For a helix $\mathbf{r}(t) = (\cos t,\, \sin t,\, t/4)$:

Here $\mathbf{r}'(t)=(-\sin t, \cos t, 1/4)$ and $|\mathbf{r}'(t)|=\sqrt{1+1/16}=\sqrt{17}/4$, a constant! So $L = \frac{\sqrt{17}}{4}(b-a)$.

The Speed Function

The integrand $|\mathbf{r}'(t)|$ is the speed of the parametrization. Faster sections contribute more length per unit parameter.

The curve $\mathbf{r}(t)=(t, \sin(2t))$ for $t\in[0, 2\pi]$ is colored by speed. Near the zero crossings of $\sin(2t)$, the curve has the steepest slope and $|\mathbf{r}'(t)|$ is largest. At the peaks and troughs, the curve momentarily levels off and speed is at its minimum.

Variable Speed - The Astroid

We measured the helix above - constant speed made it easy. What about a curve whose speed changes? The astroid $x = \cos^3 t$, $y = \sin^3 t$ crawls near its cusps and races along the sides.

We need the derivatives:

$x'(t) = -3\cos^2 t\sin t$, $\quad y'(t) = 3\sin^2 t\cos t$

The speed is

$$|\mathbf{r}'(t)| = \sqrt{9\cos^4 t\sin^2 t + 9\sin^4 t\cos^2 t} = 3|\sin t\cos t|\sqrt{\cos^2 t + \sin^2 t} = 3|\sin t\cos t|$$

Before integrating, predict: the astroid fits inside a unit circle (circumference $2\pi \approx 6.28$). Its cusps cut corners, so we expect $L < 2\pi$.

Over $[0, 2\pi]$ the function $|\sin t\cos t|$ repeats four times per period with period $\pi/2$. So we pull out the symmetry:

$$L = \int_0^{2\pi} 3|\sin t\cos t|\,dt = 4\int_0^{\pi/2} 3\sin t\cos t\,dt = 12\!\int_0^{\pi/2} \sin t\cos t\,dt$$

The substitution $u = \sin t$ gives $\int_0^{\pi/2}\sin t\cos t\,dt = \frac{1}{2}$, so

Just as expected - the astroid's perimeter is $6$, barely shorter than the unit circle's $2\pi \approx 6.28$.

Common trap: the absolute value $|\sin t\cos t|$ matters. Without it, positive and negative pieces cancel and you get $0$ - clearly wrong for a length.

Surface Area - The Idea

For a parametric surface $\mathbf{r}(u,v)$, a small parameter rectangle $\Delta u \times \Delta v$ maps to a parallelogram on the surface spanned by $\mathbf{r}_u\,\Delta u$ and $\mathbf{r}_v\,\Delta v$.

The area of that parallelogram is $|\mathbf{r}_u \times \mathbf{r}_v|\,\Delta u\,\Delta v$. Sum over all patches and take the limit:

Surface Area - Heatmap

The factor $|\mathbf{r}_u \times \mathbf{r}_v|$ varies across the surface. Brighter regions have more area per unit parameter.

Surface Area - Paraboloid from Scratch

The visualizer above shows surface areas computed for you. Let's get our hands dirty and compute one from scratch - the bowl $z = x^2 + y^2$ over the unit disk $x^2 + y^2 \le 1$.

Step 1 - Parametrize in polar coordinates.

$$\mathbf{r}(r,\theta) = r\cos\theta\,\mathbf{i} + r\sin\theta\,\mathbf{j} + r^2\,\mathbf{k}, \quad 0 \le r \le 1,\; 0 \le \theta \le 2\pi$$

Step 2 - Partial derivatives.

$\mathbf{r}_r = (\cos\theta,\; \sin\theta,\; 2r)$, $\quad \mathbf{r}_\theta = (-r\sin\theta,\; r\cos\theta,\; 0)$

Step 3 - Cross product.

$$\mathbf{r}_r \times \mathbf{r}_\theta = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & \sin\theta & 2r \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix}$$

$\mathbf{i}$-component: $\sin\theta \cdot 0 - 2r \cdot r\cos\theta = -2r^2\cos\theta$

$\mathbf{j}$-component: $-(cos\theta \cdot 0 - 2r\cdot(-r\sin\theta)) = -2r^2\sin\theta$

$\mathbf{k}$-component: $\cos\theta \cdot r\cos\theta - \sin\theta\cdot(-r\sin\theta) = r$

So $\mathbf{r}_r \times \mathbf{r}_\theta = (-2r^2\cos\theta,\; -2r^2\sin\theta,\; r)$, and its magnitude is

$$|\mathbf{r}_r \times \mathbf{r}_\theta| = \sqrt{4r^4\cos^2\theta + 4r^4\sin^2\theta + r^2} = r\sqrt{4r^2 + 1}$$

Common trap: don't forget the $r$ factor. It is part of $|\mathbf{r}_r \times \mathbf{r}_\theta|$, not a separate Jacobian you tack on afterwards.

Step 4 - Integrate.

$$S = \int_0^{2\pi}\!\int_0^1 r\sqrt{4r^2+1}\,dr\,d\theta = 2\pi \int_0^1 r\sqrt{4r^2+1}\,dr$$

Substitute $w = 4r^2 + 1$, $dw = 8r\,dr$:

$$\int_0^1 r\sqrt{4r^2+1}\,dr = \frac{1}{8}\int_1^5 \sqrt{w}\,dw = \frac{1}{8}\cdot\frac{2}{3}\bigl[w^{3/2}\bigr]_1^5 = \frac{1}{12}(5\sqrt{5} - 1)$$

Sanity check: the flat unit disk has area $\pi \approx 3.14$. The bowl bulges upward, so $S > \pi$. We got $\frac{\pi}{6}(5\sqrt{5}-1) \approx 5.33$ - significantly more, which makes sense.

Final Review - Picking the Right Integral

Arc length and surface area share a single idea: chop the curve or surface into infinitesimal pieces, measure each piece, and sum. The work on exam day is choosing the right form of the piece. Let's lay them all out side by side.

Decision flow: graph? use the $\sqrt{1+f_x^2+f_y^2}$ form. Revolution? use $\int 2\pi y\,ds$. Parametrized differently? fall back to the cross product. Almost every exam problem in §4.7 reduces to one of these.

Traps to Watch For

Surface and arc-length integrals look forgiving until you compute one and get a negative number. The errors below explain almost all of those moments.

1. Confusing $dt$ with $ds$. The differential element on a curve is $ds = |\mathbf{r}'(t)|\,dt$, not $dt$. Forgetting the $|\mathbf{r}'(t)|$ factor is the single most common slip - we end up integrating the parameter, not the arc length.
2. Using the graph formula on a non-graph. A sphere isn't $z = f(x,y)$ globally - it has two sheets. A cone has a vertex where partials blow up. When the surface fails the vertical line test, parametrize and use the cross product.
3. Sign of the cross product. Surface area uses $|\mathbf{r}_u \times \mathbf{r}_v|$ - the magnitude. Swapping $u$ and $v$ flips the cross product but the magnitude is the same, so plain surface-area answers are orientation-blind. (Flux integrals are not - that's a different story.)
4. Forgetting the $2\pi$ in surfaces of revolution. Each point on the curve traces a circle of radius (distance to axis); that circle has circumference $2\pi r$, not $r$. Without the $2\pi$ we've computed length, not area.
5. Integrating against the wrong bounds. When we parametrize, the bounds belong to the parameters $u, v$, not the original $x, y$. Mixing them yields garbage. Sketch the parameter region $R$ before integrating.
6. Pulling the speed inside a square root. $|\mathbf{r}'(t)| = \sqrt{x'^2 + y'^2 + z'^2}$ - the components are squared individually before adding, never the other way around. This trips us up on circle parametrizations where $a\cos\theta$ and $a\sin\theta$ tempt us to "factor out $a$" mid-square-root.