Measure a curved surface in three dimensions, integrals from easy to advanced lesson

Rotate the curve $y = \sqrt{x}$ , $1 \le x \le 4$ , about the $x$ -axis. It sweeps out an open, widening tube whose radius grows from $1$ to $2$ . How much area does that curved surface have, leaving out the two flat end disks and the volume inside? This page builds the surface from thin bands, each close to a slice of a cone, adds their areas with an integral and finds the exact area $\tfrac{\pi}{6}(17\sqrt{17} - 5\sqrt5) \approx 30.85$ square units.

Key terms

Lateral surface

Definition When a plane curve is rotated about a line in its plane, the lateral surface is the curved surface that the curve sweeps out. It does not include the flat disks, called end caps, that would close the surface at its boundary circles, and it is a surface measured in square units, not the solid it encloses. For a cylinder, a cone or a frustum of a cone, the lateral surface is the side without the base or bases.

In plain words Think of a lampshade rather than a closed can. The lateral surface is the shade itself, and its openings at the ends stay open.

Example A cylinder of radius $r$ and height $h$ has lateral area $2\pi r h$ , the area of the rectangle you get by cutting its side along a line parallel to the axis and unrolling it. Its total surface area, with both end caps, is $2\pi r h + 2\pi r^2$ . For $y = \sqrt{x}$ on $[1, 4]$ rotated about the $x$ -axis, the boundary circles have radius $1$ at $x = 1$ and radius $2$ at $x = 4$ . End caps would add $\pi + 4\pi = 5\pi$ , and they are not part of the lateral surface. The solid inside has volume $\pi \int_1^4 x\,dx = \tfrac{15\pi}{2}$ cubic units, which is a different quantity again.

Conical frustum

Definition A conical frustum is the part of a right circular cone between two planes perpendicular to its axis. Let its boundary circles have radii $r_1$ and $r_2$ , and let its slant length $\ell$ be the length of a straight segment on the cone that joins the two circles. Then its lateral area is

\begin{gathered} A = \pi (r_1 + r_2)\,\ell = 2\pi \bar{r}\,\ell, \\[0.4em] \bar{r} = \frac{r_1 + r_2}{2}. \end{gathered}

The formula comes from the cone. A cone with base radius $r$ and slant length $s$ unrolls into a circular sector of radius $s$ and arc length $2\pi r$ , whose area is $\tfrac12 \cdot s \cdot 2\pi r = \pi r s$ . A frustum is a large cone with a similar small cone removed, and subtracting the two lateral areas gives $\pi (r_1 + r_2)\,\ell$ . The same formula gives the lateral area of a cylinder when $r_1 = r_2$ and of a cone when $r_1 = 0$ .

In plain words Cut the side of a frustum open and flatten it, and you get part of a flat ring. Its area is the average of the two boundary circumferences times the slant width.

Example Rotating the chord from $(1, 1)$ to $(4, 2)$ about the $x$ -axis gives a frustum with $r_1 = 1$ , $r_2 = 2$ and slant length $\ell = \sqrt{3^2 + 1^2} = \sqrt{10}$ . Its lateral area is $3\pi\sqrt{10} \approx 29.80$ . This frustum is a first approximation of the surface swept by $y = \sqrt{x}$ . The chords over three equal pieces of $[1, 4]$ give three frustums with total area about $30.72$ , and six pieces give about $30.81$ .

Slant length

Definition The slant length of a frustum with radii $r_1$ and $r_2$ whose planes are a distance $h$ apart is

\ell = \sqrt{h^2 + (r_2 - r_1)^2},

the Pythagorean length of the segment that sweeps it out. For a surface of revolution, the slant length of the thin band swept by the piece of the curve over $[x, x + \Delta x]$ is the arc length of that piece. As $\Delta x$ shrinks it becomes the arc length element

ds = \sqrt{1 + f'(x)^2}\,dx,

which is never less than the width $dx$ measured along the axis, and equals it only where the tangent is parallel to the axis.

In plain words Slant length is measured along the surface, following its slope, not along the axis. A band on a steep part of the surface is wider than its shadow on the axis.

Example For $y = \sqrt{x}$ the slant factor is $\sqrt{1 + \tfrac{1}{4x}}$ , which is $\tfrac{\sqrt5}{2} \approx 1.118$ at $x = 1$ and $\tfrac{\sqrt{17}}{4} \approx 1.031$ at $x = 4$ . The whole generating curve from $(1, 1)$ to $(4, 2)$ has length

\int_1^4 \sqrt{1 + \frac{1}{4x}}\,dx \approx 3.1678,

a little more than the chord length $\sqrt{10} \approx 3.1623$ and more than the axial length $3$ .

Surface area element

Definition A short piece of a curve with arc length element $ds$ , at distance $r$ from the axis of rotation, sweeps out a thin band with area element

dS = 2\pi r\,ds.

For the graph of a function $f$ with $f(x) \ge 0$ , rotated about the $x$ -axis, the radius is $r = f(x)$ and

\begin{aligned} dS &= 2\pi f(x)\,ds \\[0.4em] &= 2\pi f(x)\sqrt{1 + f'(x)^2}\,dx. \end{aligned}

In plain words The band is a narrow strip that goes once around the axis. Its length is the circumference $2\pi r$ , its width is the slant width $ds$ , and the two directions are perpendicular. Using $dx$ in place of $ds$ treats every band as a short cylinder parallel to the axis, and it loses area wherever the curve slopes.

Example For $y = \sqrt{x}$ on $[1, 4]$ the derivative is $y' = \tfrac{1}{2\sqrt{x}}$ , so

dS = 2\pi\sqrt{x}\,\sqrt{1 + \tfrac{1}{4x}}\,dx.

At $x = 1$ this is $\pi\sqrt5\,dx$ , about $7.02\,dx$ , and at $x = 4$ it is $\pi\sqrt{17}\,dx$ , about $12.95\,dx$ . Replacing $ds$ by $dx$ would give $2\pi \int_1^4 \sqrt{x}\,dx = \tfrac{28\pi}{3} \approx 29.32$ , less than the true area found below. The volume elements of the washer method and the shell method, $\pi (R^2 - r^2)\,dx$ and $2\pi r h\,dr$ , are measured in cubic units, while $2\pi r\,ds$ is measured in square units.

Surface area of revolution

Definition Let $f$ be continuously differentiable on $[a, b]$ with $a < b$ , and let $f(x) \ge 0$ for every $x$ in $[a, b]$ . Rotating the graph of $y = f(x)$ , $a \le x \le b$ , about the $x$ -axis gives a lateral surface whose area is

S = 2\pi \int_a^b y \sqrt{1 + (y')^2}\,dx.

This area is defined as the limit of the total lateral area of the frustums built on inscribed chords, as the widest piece shrinks. By the mean value theorem, the frustum over $[x_{i-1}, x_i]$ has area $\pi\bigl(f(x_{i-1}) + f(x_i)\bigr)\sqrt{1 + f'(c_i)^2}\,\Delta x_i$ for some $c_i$ in $(x_{i-1}, x_i)$ . Because $f$ is continuous and $f'$ is bounded, these sums tend to the integral. If instead $a \ge 0$ and the graph is rotated about the $y$ -axis, the radius is $x$ and the area is $2\pi \int_a^b x \sqrt{1 + f'(x)^2}\,dx$ . End caps are never included.

In plain words Each band contributes its circumference times its slant width, and the integral adds the bands from one end of the surface to the other. The radius is the distance from the curve to the axis, which for rotation about the $x$ -axis is the height $f(x)$ .

Example Rotate $y = \sqrt{x}$ , $1 \le x \le 4$ , about the $x$ -axis, keeping only the lateral surface. On $[1, 4]$ the function is positive and its derivative $\tfrac{1}{2\sqrt{x}}$ is continuous, so

S = 2\pi \int_1^4 \sqrt{x}\,\sqrt{1 + \tfrac{1}{4x}}\,dx.

Both radicands are positive for $x > 0$ , so the square roots can be combined,

\begin{aligned} \sqrt{x}\,\sqrt{1 + \tfrac{1}{4x}} &= \sqrt{x + \tfrac14} \\[0.4em] &= \tfrac12 \sqrt{4x + 1}. \end{aligned}

The factor $\tfrac12$ cancels the $2$ in front. An antiderivative of $\sqrt{4x + 1}$ is $\tfrac16 (4x + 1)^{3/2}$ , since its derivative is $\tfrac16 \cdot \tfrac32 \cdot 4 \sqrt{4x + 1} = \sqrt{4x + 1}$ . Hence

\begin{aligned} S &= \pi \int_1^4 \sqrt{4x + 1}\,dx \\[0.4em] &= \frac{\pi}{6}\Bigl[(4x + 1)^{3/2}\Bigr]_1^4 \\[0.4em] &= \frac{\pi}{6}\bigl(17\sqrt{17} - 5\sqrt{5}\bigr) \\[0.4em] &\approx 30.85 \end{aligned}

square units. The integrand $\pi\sqrt{4x + 1}$ increases from $\pi\sqrt5$ to $\pi\sqrt{17}$ over an interval of length $3$ , so the area lies between $3\pi\sqrt5 \approx 21.07$ and $3\pi\sqrt{17} \approx 38.86$ . The frustum totals $29.80$ , $30.72$ and $30.81$ approach it from below.

Check yourself

Rotate the segment $y = \tfrac{x}{2} + 1$ , $0 \le x \le 4$ , about the $x$ -axis. Find the lateral surface area with the integral and check it with the frustum formula. What is the total area if both end disks are included?

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Here $y' = \tfrac12$ , so $\sqrt{1 + (y')^2} = \tfrac{\sqrt5}{2}$ and

\begin{aligned} S &= 2\pi \int_0^4 \left(\tfrac{x}{2} + 1\right)\tfrac{\sqrt5}{2}\,dx \\[0.4em] &= \pi\sqrt5 \Bigl[\tfrac{x^2}{4} + x\Bigr]_0^4 \\[0.4em] &= 8\pi\sqrt5 \approx 56.20. \end{aligned}

The surface is a frustum with radii $1$ and $3$ and slant length $\sqrt{4^2 + 2^2} = 2\sqrt5$ , and the frustum formula gives $\pi (1 + 3) \cdot 2\sqrt5 = 8\pi\sqrt5$ , in agreement. The end disks have areas $\pi$ and $9\pi$ , so the total is $8\pi\sqrt5 + 10\pi \approx 87.61$ square units. The integral alone never includes them.

Find the area of the surface swept out when $y = x^3$ , $0 \le x \le 1$ , is rotated about the $x$ -axis.

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The function is nonnegative and continuously differentiable on $[0, 1]$ , with $y' = 3x^2$ , so

S = 2\pi \int_0^1 x^3 \sqrt{1 + 9x^4}\,dx.

Substitute $u = 1 + 9x^4$ . Then $du = 36x^3\,dx$ , and the bounds $x = 0$ and $x = 1$ become $u = 1$ and $u = 10$ , so

\begin{aligned} S &= \frac{2\pi}{36} \int_1^{10} u^{1/2}\,du \\[0.4em] &= \frac{\pi}{18} \cdot \frac23 \bigl(10^{3/2} - 1\bigr) \\[0.4em] &= \frac{\pi}{27}\bigl(10\sqrt{10} - 1\bigr) \\[0.4em] &\approx 3.563. \end{aligned}

Using $dx$ in place of $ds$ would give only $2\pi \int_0^1 x^3\,dx = \tfrac{\pi}{2} \approx 1.571$ , far too small, because the curve is steep near $x = 1$ .

Rotating the semicircle $y = \sqrt{25 - x^2}$ about the $x$ -axis gives a sphere of radius $5$ . Find the area of the part of the sphere with $-3 \le x \le 4$ . Then explain why the surface area formula does not apply directly on $[-5, 5]$ , and find the area of the whole sphere anyway.

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For $-5 < x < 5$ the derivative is $y' = -\tfrac{x}{\sqrt{25 - x^2}}$ , so

\begin{aligned} 1 + (y')^2 &= \frac{25}{25 - x^2}, \\[0.4em] y \sqrt{1 + (y')^2} &= 5. \end{aligned}

The function is continuously differentiable on $[-3, 4]$ , so the part with $-3 \le x \le 4$ has area

\begin{aligned} S &= 2\pi \int_{-3}^{4} 5\,dx \\[0.4em] &= 70\pi \approx 219.9. \end{aligned}

At $x = \pm 5$ the derivative does not exist, and $\lvert y' \rvert$ grows without bound near those points, so the function is not continuously differentiable on $[-5, 5]$ . Apply the formula on $[-t, t]$ for $0 < t < 5$ instead, which gives $2\pi \cdot 5 \cdot 2t = 20\pi t$ , and let $t \to 5^-$ . These zones fill out the sphere, so its area is the limit $100\pi$ , which matches $4\pi r^2$ with $r = 5$ .