Arc Length and Surface Area using Integration

Where the formula comes from. Zoom in on a tiny piece of a curve. Over a small step the curve is almost straight, so the little arc length $\delta s$ is the hypotenuse of a right-angled triangle with horizontal leg $\delta x$ and vertical leg $\delta y$. By Pythagoras:

$\delta s \approx \sqrt{(\delta x)^2 + (\delta y)^2}.$

Factor out $\delta x$ to bring in the gradient: $\delta s \approx \sqrt{1 + \left(\frac{\delta y}{\delta x}\right)^2}\;\delta x.$

Summing these elements and taking the limit $\delta x \to 0$ turns $\frac{\delta y}{\delta x}$ into $\frac{dy}{dx}$ and the sum into an integral — the Cartesian arc-length formula: $s = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\;dx.$

The same triangle, factored differently, gives the two forms you will use:

• factor out $\delta x \Rightarrow ds = \sqrt{1+(\tfrac{dy}{dx})^2}\,dx$ (Cartesian);
• factor out $\delta t \Rightarrow ds = \sqrt{(\tfrac{dx}{dt})^2 + (\tfrac{dy}{dt})^2}\,dt$ (parametric).

Cambridge tip. Knowing $ds$ comes from one small triangle is what lets you adapt to any setting — Cartesian, parametric, or surface area — without memorising four separate things.

For a curve $y=f(x)$ between $x=a$ and $x=b$, the routine is mechanical:

1. Differentiate to get $\dfrac{dy}{dx}$.
2. Square it and add $1$: form $1 + \left(\dfrac{dy}{dx}\right)^2$.
3. Simplify — exam curves are engineered so this becomes a perfect square that the root removes cleanly.
4. Integrate the simplified root between the limits.

Where the magic comes from. A favourite design uses $y = \cosh x$ (a catenary). Then $\dfrac{dy}{dx} = \sinh x$, and the identity $1 + \sinh^2 x = \cosh^2 x$ makes the root collapse beautifully: $\sqrt{1 + \sinh^2 x} = \sqrt{\cosh^2 x} = \cosh x.$ So $s = \int_a^b \cosh x\,dx = \big[\sinh x\big]_a^b$ — a one-line integral.

Algebraic curves work the same way. For $y = \dfrac{x^2}{4} - \dfrac{\ln x}{2}$ type curves, $1+(\frac{dy}{dx})^2$ factors into $\left(\frac{dy}{dx}+\text{something}\right)^2$. Always look for a perfect square before reaching for a substitution — if it does not simplify, re-check your differentiation.

Cambridge tip. If $1+(\frac{dy}{dx})^2$ refuses to become a perfect square, you have almost certainly made a slip differentiating. The whole point of the set question is that the root disappears.

When a curve is given by $x = x(t)$, $y = y(t)$, factor the arc element by $\delta t$ instead of $\delta x$: $s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;dt.$

The routine:

1. Differentiate both $x$ and $y$ with respect to $t$.
2. Square and add: form $\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2$.
3. Simplify (again, engineered to be a perfect square) and integrate over the $t$-range.

Classic example — a circle $x = r\cos t$, $y = r\sin t$: $\frac{dx}{dt} = -r\sin t,\quad \frac{dy}{dt} = r\cos t,$ $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = r^2\sin^2 t + r^2\cos^2 t = r^2.$ So $s = \int_0^{2\pi} r\,dt = 2\pi r$ — the circumference, exactly as it should be. This sanity-check ($\sin^2+\cos^2=1$) is the parametric cousin of the $1+\sinh^2=\cosh^2$ trick.

Cambridge tip. Make sure the limits are values of $t$, not of $x$ or $y$. Converting the given $x$- or $y$-endpoints into the corresponding $t$-values is a step candidates routinely forget.

The picture. Rotate the arc about the $x$-axis. A tiny arc element of length $ds$, sitting at height $y$ above the axis, sweeps out a thin band (a frustum ribbon). Its circumference is $2\pi y$ (a circle of radius $y$) and its slant width is $ds$, so its area is $2\pi y\,ds$. Summing all the ribbons:

$S = \int 2\pi y\,ds.$

Substitute the right $ds$. Replace $ds$ with whichever form fits the curve:

• Cartesian: $S = \displaystyle\int_a^b 2\pi y\,\sqrt{1+\left(\tfrac{dy}{dx}\right)^2}\,dx$;
• Parametric: $S = \displaystyle\int_{t_1}^{t_2} 2\pi y\,\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\,dt$.

About the $y$-axis instead? The radius of each band is now the horizontal distance $x$, so $S = \int 2\pi x\,ds.$ Choose the radius that matches the spin axis: $y$ for rotation about the $x$-axis, $x$ for rotation about the $y$-axis.

Cambridge tip. Two things must agree before integrating: the radius ($y$ or $x$) must match the axis of rotation, and the $ds$ must be expressed in the variable you are integrating with respect to. Mixing $x\,ds$ with $dx$ when the spin is about the $x$-axis is the standard error.