2.4 Arc Length of a Curve and Surface Area

Big idea

So far the slicing procedure has measured regions and solids. The same procedure measures a curve. Cut the curve into short pieces, approximate one piece by the straight segment joining its endpoints, add the lengths, and let the pieces shrink. The only thing you need is the length of one small segment, and that is the distance formula.

Over a short run $\Delta x$, a differentiable function rises by about $f'(x)\,\Delta x$. The segment from one endpoint to the next is the hypotenuse of a right triangle with legs $\Delta x$ and $f'(x)\Delta x$, so its length is $\sqrt{(\Delta x)^2 + (f'(x)\Delta x)^2} = \sqrt{1 + [f'(x)]^2}\,\Delta x$. Adding these and passing to the limit gives the arc length integral. The factor $\sqrt{1 + [f'(x)]^2}$ is worth naming: it is how much longer a piece of curve is than its shadow on the horizontal axis. Where the graph is flat the factor is $1$; where the graph is steep the factor is large.

Surface area of a solid of revolution comes from the same segment. Rotate that short piece about the $x$-axis and it sweeps out a thin band. Because the piece is straight, the band is the lateral surface of a slice of a cone, whose area is the circumference at its average radius times its slant length, so it contributes $2\pi f(x)\sqrt{1 + [f'(x)]^2}\,dx$. The slant, not the horizontal run, is the correct factor: using $dx$ there would systematically undercount, and it is why a cone’s lateral area involves its slant height rather than its height.

A practical catch shapes how these problems are written. The integrand carries a square root of a sum, and for most functions that root has no elementary antiderivative; even a plain parabola produces an integral you cannot finish by hand. So the examples you meet are engineered so that $1 + [f'(x)]^2$ collapses into a perfect square. Recognizing that collapse is the skill being tested; elsewhere you set the integral up and hand it to a numerical method.

Decoder

If $f$ has a continuous derivative on $[a, b]$, the length of the graph of $f$ over that interval is the integral from $a$ to $b$ of the square root of $1$ plus the square of $f'(x)$.

The hypothesis is that $f'$ is continuous, not merely that it exists. A curve with a corner has a jump in $f'$, and while you can still measure it by splitting at the corner, a derivative that misbehaves badly enough can leave the curve with no finite length at all.

The expression $ds = \sqrt{1 + [f'(x)]^2}\,dx$ is the arc length element, and both formulas here are built from it. Length is $\int ds$; surface area of a revolution is $\int 2\pi r\, ds$, where $r$ is the distance from the axis to the curve. Writing them that way keeps two facts visible: the radius belongs to the axis you are rotating about, and the thickness factor is always the slant $ds$, never $dx$.

The standard slip is the missing $1$. The integrand is $\sqrt{1 + [f'(x)]^2}$, not $\sqrt{[f'(x)]^2}$, and not $1 + f'(x)$. A fast audit: for the horizontal line $f(x) = c$ the integrand must reduce to $1$, giving length $b - a$. Any formula that fails that test is wrong.

Definitions and results

The arc length element. For a differentiable function,

$$ ds = \sqrt{1 + \big[ f'(x) \big]^2}\; dx $$

This is the length of the piece of curve lying above a run of width $dx$.

Arc length. If $f'$ is continuous on $[a, b]$, the length of the graph is

$$ L = \int_a^b \sqrt{1 + \big[ f'(x) \big]^2}\; dx $$

Arc length in terms of $y$. If the curve is given as $x = g(y)$ with $g'$ continuous on $[c, d]$, slice horizontally instead:

$$ L = \int_c^d \sqrt{1 + \big[ g'(y) \big]^2}\; dy $$

Use whichever description avoids solving for the other variable, and note that a curve with a vertical tangent, such as $x = y^2$ near the origin, is far easier to handle in this form.

Surface of revolution about the $x$-axis. If $f(x) \ge 0$ and $f'$ is continuous on $[a, b]$, rotating the graph about the $x$-axis produces a surface of area

$$ S = 2\pi \int_a^b f(x) \sqrt{1 + \big[ f'(x) \big]^2}\; dx $$

Surface of revolution about the $y$-axis. With the curve written as $x = g(y) \ge 0$ on $[c, d]$,

$$ S = 2\pi \int_c^d g(y) \sqrt{1 + \big[ g'(y) \big]^2}\; dy $$

In both statements the factor multiplying $2\pi$ is the distance from the axis of rotation to the curve.

Why the slant appears. A short straight piece at distance $r$ from the axis sweeps a frustum whose lateral area is $2\pi r \ell$, with $\ell$ the slant length. Replacing $\ell$ by the horizontal run would shrink every band and give an answer that is too small, by a factor that grows with the slope.

Integrands that simplify. The useful cases are those where $1 + [f'(x)]^2$ is a perfect square. Two patterns do this reliably. The first is a power such as $f(x) = c\,x^{3/2}$, whose derivative squares to a linear function, leaving a single easy radical. The second is any $f$ whose derivative has the shape

$$ f'(x) = \frac{1}{2}\left( w - \frac{1}{w} \right) \quad \Longrightarrow \quad 1 + \big[ f'(x) \big]^2 = \frac{1}{4}\left( w + \frac{1}{w} \right)^2 $$

for some expression $w$, since the cross terms are $-\tfrac{1}{2}$ and $+\tfrac{1}{2}$ respectively. The radical then disappears entirely.

Everything else. Most arc length integrals, including the one for $y = x^2$, have no elementary antiderivative. Setting up the correct integral is a complete answer to that kind of problem.

Worked examples

A power that flattens the radical

Find the length of $y = \tfrac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$.

Differentiate: $f'(x) = \tfrac{2}{3} \cdot \tfrac{3}{2} x^{1/2} = x^{1/2}$. Then $[f'(x)]^2 = x$, so the radical becomes $\sqrt{1 + x}$, with no square root of a square root left over.

$$ L = \int_0^3 \sqrt{1 + x}\; dx = \left[ \frac{2}{3}(1 + x)^{3/2} \right]_0^3 = \frac{2}{3}\big( 8 - 1 \big) = \frac{14}{3} $$

Check the size. The curve runs from $(0, 0)$ to $(3, 2\sqrt{3}) \approx (3, 3.46)$, and the straight line between those points has length $\sqrt{9 + 12} = \sqrt{21} \approx 4.58$. The curve must be longer than the chord, and $\tfrac{14}{3} \approx 4.67$ is longer by a little, which matches a graph that bends only gently.

A sum engineered to be a perfect square

Find the length of $y = \tfrac{x^3}{6} + \tfrac{1}{2x}$ from $x = 1$ to $x = 2$.

The derivative is $f'(x) = \tfrac{x^2}{2} - \tfrac{1}{2x^2}$. Square it:

$$ \big[ f'(x) \big]^2 = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4} $$

The cross term is exactly $-\tfrac{1}{2}$, so adding $1$ flips its sign:

$$ 1 + \big[ f'(x) \big]^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left( \frac{x^2}{2} + \frac{1}{2x^2} \right)^2 $$

On $[1, 2]$ that inner expression is positive, so its square root is itself:

$$ L = \int_1^2 \left( \frac{x^2}{2} + \frac{1}{2x^2} \right) dx = \left[ \frac{x^3}{6} - \frac{1}{2x} \right]_1^2 = \left( \frac{4}{3} - \frac{1}{4} \right) - \left( \frac{1}{6} - \frac{1}{2} \right) = \frac{13}{12} + \frac{4}{12} = \frac{17}{12} $$

The antiderivative differing from the original function only in one sign is the signature of this family; if your integrand does not collapse this way, recheck the algebra before assuming the problem is unsolvable.

A surface of revolution

The curve $y = \sqrt{x}$ for $0 \le x \le 2$ is rotated about the $x$-axis. Find the area of the surface.

Here $f'(x) = \tfrac{1}{2\sqrt{x}}$, so $1 + [f'(x)]^2 = 1 + \tfrac{1}{4x}$. Multiplying by $f(x) = \sqrt{x}$ before taking the root is what makes this manageable:

$$ \sqrt{x}\,\sqrt{1 + \frac{1}{4x}} = \sqrt{x\left(1 + \frac{1}{4x}\right)} = \sqrt{x + \frac{1}{4}} $$

The trouble at $x = 0$ disappears under that combination.

$$ S = 2\pi \int_0^2 \sqrt{x + \frac{1}{4}}\; dx = 2\pi \left[ \frac{2}{3}\left( x + \frac{1}{4} \right)^{3/2} \right]_0^2 = \frac{4\pi}{3}\left[ \left(\frac{9}{4}\right)^{3/2} - \left(\frac{1}{4}\right)^{3/2} \right] $$

Now $\left(\tfrac{9}{4}\right)^{3/2} = \left(\tfrac{3}{2}\right)^3 = \tfrac{27}{8}$ and $\left(\tfrac{1}{4}\right)^{3/2} = \left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$, so the bracket is $\tfrac{26}{8} = \tfrac{13}{4}$ and

$$ S = \frac{4\pi}{3} \cdot \frac{13}{4} = \frac{13\pi}{3} $$

A cone, checked against geometry

Rotate the segment $y = 2x$ for $0 \le x \le 1$ about the $x$-axis and find the surface area.

Here $f'(x) = 2$, so the radical is the constant $\sqrt{5}$:

$$ S = 2\pi \int_0^1 2x \sqrt{5}\; dx = 2\pi \sqrt{5} \left[ x^2 \right]_0^1 = 2\sqrt{5}\,\pi $$

This surface is a cone of base radius $2$ and height $1$, whose slant height is $\sqrt{2^2 + 1^2} = \sqrt{5}$. The lateral area of a cone is $\pi r \ell = \pi \cdot 2 \cdot \sqrt{5}$, matching exactly. Had the formula used $dx$ instead of the slant element, the answer would have come out $2\pi$, too small by a factor of $\sqrt{5}$, which is precisely the error the slant factor prevents.

Practice

Each problem gives a curve and an interval. Differentiate, form $1 + [f'(x)]^2$, and look hard for a perfect square before integrating; if the curve is easier to describe with $x$ as a function of $y$, set the integral up that way instead. The surface area problems add a radius factor, so name the axis of rotation first.

Quiz

Five items on the arc length element, lengths of curves in either variable, and surfaces of revolution.