Surface Area of Revolution: Formula and Examples

What Is Surface Area Of Revolution?

Surface area of revolution is the area of the curved skin you get when a curve is rotated a full turn around a fixed line, called the axis of revolution. Each point on the curve travels around a circle, so the surface is built from circular bands. For a curve $y = f(x)$ from $x = a$ to $x = b$, rotated about the x-axis, the area is

$$S = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}},dx$$

Here is what each symbol means:

• $S$ is the area of the curved surface, in square units.

• $y = f(x)$ is the height of the curve, which is also the radius of the circle each point travels around the x-axis.

• $2\pi y$ is the circumference of that circle.

• $\sqrt{1 + \left(\frac{dy}{dx}\right)^{2}},dx$ is the arc-length element $ds$, the true length of a tiny piece of the curve. It comes straight from the arc length of a curve.

• $a$ and $b$ are the x-values where the curve starts and stops.

The formula needs two conditions. The function must be smooth on $[a, b]$, meaning $\frac{dy}{dx}$ exists and is continuous there, and $y \geq 0$ so that $y$ really is a distance. The result is the lateral surface only: the flat circular ends of a solid are not included unless a problem asks for the total surface.

In words: surface area of revolution equals circumference times arc length, added along the curve. Keep that sentence, because every version of the formula below is the same idea with a different radius or a different $ds$.

How Is The Surface Area Of Revolution Formula Derived?

The surface area of revolution formula comes from chopping the curve into short straight segments and rotating each one. A rotated straight segment is not a cylinder. It is a frustum, a cone with its tip cut off, and its slanted side is the piece of surface we need.

Start with a full cone of base radius $r$ and slant height $l$. Cut along the slant and its side flattens into a sector of a circle of radius $l$, whose arc length is $2\pi r$. That sector has area $\frac{1}{2}(2\pi r)(l) = \pi r l$, which answers a common question about the lateral surface area of a cone. Removing a smaller cone from the top leaves a frustum with end radii $r_1$ and $r_2$ and slant length $L$, and its side area is

$$A_{\text{frustum}} = \pi (r_1 + r_2) L = 2\pi \left(\frac{r_1 + r_2}{2}\right) L$$

So a frustum band has area equal to the circumference at its average radius times its slant length. Now apply this to the curve:

1. Split $[a, b]$ into $n$ pieces of width $\Delta x$, with points $x_0, x_1, \ldots, x_n$.

2. Join neighbouring points on the curve with straight segments. By the distance formula, a segment has length $\sqrt{(\Delta x)^2 + (\Delta y)^2}$.

3. By the mean value theorem, some point $x_i^{*}$ in each piece has slope $\frac{dy}{dx}$ equal to $\frac{\Delta y}{\Delta x}$, so the segment length becomes $\sqrt{1 + \left(\frac{dy}{dx}\right)^{2}},\Delta x$ at that point.

4. Rotating the segment gives a frustum whose average radius tends to $f(x_i)$ as the pieces shrink.

5. Adding the bands gives a Riemann sum, and as $n \to \infty$ it becomes the definite integral in the formula above.

Why Doesn't Rotating Thin Cylinders Work?

Thin cylinders, which add up $2\pi y,dx$, fail because $dx$ measures only the horizontal width of a band, while the surface stretches along the slanted curve. A Physics Forums thread asks exactly this: "Surface area of revolution: why use arc length instead of dx". Volumes get away with $dx$ in the disk method, so the question is fair.

The answer lies in how big the error is.

A thin disk and a thin frustum hold almost the same volume. The gap between them shrinks like $(\Delta x)^2$, so it vanishes when the slices are added up, which is why the volume of solids of revolution only needs $dx$. Their side areas, though, differ by a factor of about $\sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}$ in every slice. That error never shrinks relative to the band itself, so it never goes away.

This distinction is subtle enough that it takes a second look at the error terms to see why one shape works for volume and fails for area. Example 2 below shows the size of the damage on a sphere.

What Is The Formula For Revolution About The Y-Axis?

Rotating about the y-axis changes the radius: each point now circles the y-axis, so its radius is its horizontal distance $x$. The arc-length element stays the same piece of curve, written in whichever variable is convenient.

If the curve is given as $y = f(x)$ with $x \geq 0$, integrate in $x$:

$$S = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}},dx$$

If the curve is given as $x = g(y)$ from $y = c$ to $y = d$, integrate in $y$:

$$S = 2\pi \int_{c}^{d} x \sqrt{1 + \left(\frac{dx}{dy}\right)^{2}},dy$$

A first draft of the rule would read "the radius is $x$ for the y-axis and $y$ for the x-axis." Scratch that: the radius is the distance from the point to the axis, which equals $x$ only because the y-axis is the line $x = 0$. Rotate about the line $y = -1$ and the radius becomes $y + 1$. Rotate $y = 2$ on $[0, 3]$ about $y = -1$ and you get a cylinder of radius $3$, with area $2\pi(3)(3) = 18\pi$.

What Is The Parametric Form Of Surface Area Of Revolution?

When a curve is traced by parametric equations $x = x(t)$, $y = y(t)$ for $\alpha \leq t \leq \beta$, the arc-length element becomes $ds = \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}},dt$. The surface area of revolution about each axis is

$$S_x = 2\pi \int_{\alpha}^{\beta} y(t) \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}},dt$$

$$S_y = 2\pi \int_{\alpha}^{\beta} x(t) \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}},dt$$

Two conditions matter here. The curve must be traced exactly once as $t$ runs from $\alpha$ to $\beta$, or parts of the surface get counted twice. The radius, $y(t)$ or $x(t)$, must stay non-negative on the interval.

How Does Surface Area Of Revolution Work In Polar Form?

For a polar curve $r = r(\theta)$, write the point as $x = r\cos\theta$ and $y = r\sin\theta$. The arc-length element becomes $ds = \sqrt{r^{2} + \left(\frac{dr}{d\theta}\right)^{2}},d\theta$, and rotating about the polar axis (the x-axis) gives $S = 2\pi \int y,ds$ with $y = r\sin\theta$.

A homework question on Study.com asks you to show that the cardioid $r = 1 + \cos\theta$ has a surface area of revolution of $\frac{32\pi}{5}$. The upper half of the cardioid runs from $\theta = 0$ to $\theta = \pi$.

$$r^{2} + \left(\frac{dr}{d\theta}\right)^{2} = (1 + \cos\theta)^{2} + \sin^{2}\theta = 2 + 2\cos\theta = 4\cos^{2}\frac{\theta}{2}$$

$$S = 2\pi \int_{0}^{\pi} (1 + \cos\theta)\sin\theta \cdot 2\cos\frac{\theta}{2},d\theta$$

$$S = 2\pi \int_{0}^{\pi} 8\cos^{4}\frac{\theta}{2}\sin\frac{\theta}{2},d\theta$$

With $u = \cos\frac{\theta}{2}$, the integral becomes $16\int_{0}^{1} u^{4},du = \frac{16}{5}$, so $S = \frac{32\pi}{5}$, as the question claims.

Which Surface Area Of Revolution Formula Should You Use?

Every formula is $2\pi \times (\text{radius}) \times ds$. Only the radius and the form of $ds$ change.

Pick the variable that makes $ds$ simplest, not the one the problem happened to use. A curve like $x = \frac{1}{3}y^{3}$ is much easier in $y$ than solved for $y$ in terms of $x$.

Examples Of Surface Area Of Revolution

Example 1

Find the surface area of revolution when $y = \frac{3}{4}x$, $0 \leq x \leq 4$, is rotated about the x-axis.

The slope is constant:

$$\frac{dy}{dx} = \frac{3}{4}$$

$$\sqrt{1 + \left(\frac{3}{4}\right)^{2}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$$

$$S = 2\pi \int_{0}^{4} \frac{3}{4}x \cdot \frac{5}{4},dx = 2\pi \cdot \frac{15}{16} \cdot 8$$

$$S = 15\pi$$

The surface is a cone with radius $3$ and slant height $5$, and the cone formula $\pi r l = \pi(3)(5) = 15\pi$ agrees.

Final answer: $15\pi \approx 47.1239$ square units.

Example 2

Find the surface area of a sphere of radius 2 by rotating $y = \sqrt{4 - x^{2}}$, $-2 \leq x \leq 2$, about the x-axis.

The tempting route treats each band as a cylinder and adds up $2\pi y,dx$:

$$S_{\text{wrong}} = 2\pi \int_{-2}^{2} \sqrt{4 - x^{2}},dx = 2\pi \cdot 2\pi = 4\pi^{2} \approx 39.4784$$

That number fails a check you already know: the sphere area formula $4\pi r^{2}$ gives $16\pi \approx 50.2655$. The cylinder bands are too short near the ends, where the curve is steep. Now use the correct $ds$:

$$\frac{dy}{dx} = \frac{-x}{\sqrt{4 - x^{2}}}$$

$$1 + \left(\frac{dy}{dx}\right)^{2} = \frac{4}{4 - x^{2}}$$

$$y\sqrt{1 + \left(\frac{dy}{dx}\right)^{2}} = \sqrt{4 - x^{2}} \cdot \frac{2}{\sqrt{4 - x^{2}}} = 2$$

$$S = 2\pi \int_{-2}^{2} 2,dx = 16\pi$$

The slope blows up at $x = \pm 2$, but the product simplifies to the constant $2$ before integrating, so nothing breaks.

Final answer: $16\pi \approx 50.2655$ square units, matching $4\pi r^{2}$.

Example 3

Find the area of the paraboloid formed by rotating $y = \sqrt{x}$, $0 \leq x \leq 2$, about the x-axis.

$$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$$

$$1 + \left(\frac{dy}{dx}\right)^{2} = \frac{4x + 1}{4x}$$

$$y\sqrt{1 + \left(\frac{dy}{dx}\right)^{2}} = \sqrt{x} \cdot \frac{\sqrt{4x + 1}}{2\sqrt{x}} = \frac{\sqrt{4x + 1}}{2}$$

$$S = \pi \int_{0}^{2} \sqrt{4x + 1},dx = \pi \left[\frac{(4x + 1)^{3/2}}{6}\right]_{0}^{2} = \frac{\pi}{6}(27 - 1)$$

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

The step that looks skippable is multiplying $y$ into the square root before integrating. Students who integrate $\sqrt{x}\sqrt{1 + \frac{1}{4x}}$ as written stall; the ones who simplify first see a plain power of $4x + 1$, which integration by substitution finishes in one line.

Final answer: $\frac{13\pi}{3} \approx 13.6136$ square units.

Example 4

Rotate $y = x^{2}$, $0 \leq x \leq 1$, about the y-axis and find the surface area.

The radius is $x$, and $\frac{dy}{dx} = 2x$:

$$S = 2\pi \int_{0}^{1} x\sqrt{1 + 4x^{2}},dx$$

Let $u = 1 + 4x^{2}$, so $du = 8x,dx$ and the limits become $1$ and $5$:

$$S = \frac{2\pi}{8} \int_{1}^{5} \sqrt{u},du = \frac{\pi}{4} \cdot \frac{2}{3}\left(5^{3/2} - 1\right)$$

$$S = \frac{\pi}{6}\left(5\sqrt{5} - 1\right)$$

Final answer: $\frac{\pi}{6}(5\sqrt{5} - 1) \approx 5.3304$ square units, the inside of a shallow bowl.

Example 5

The loop of $x = t^{2}$, $y = t - \frac{t^{3}}{3}$ is rotated about the x-axis. Find the surface area.

This is a Brainly question: prove that the surface area generated by revolving the loop about the x-axis is $3\pi$. The loop closes where $y = 0$ and $t \neq 0$, at $t = \pm\sqrt{3}$, and the upper half runs over $0 \leq t \leq \sqrt{3}$, where $y \geq 0$. The lower half sweeps out the same surface, so we rotate only the upper half.

$$\frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 1 - t^{2}$$

$$\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2} = 4t^{2} + 1 - 2t^{2} + t^{4} = \left(1 + t^{2}\right)^{2}$$

$$S = 2\pi \int_{0}^{\sqrt{3}} \left(t - \frac{t^{3}}{3}\right)\left(1 + t^{2}\right),dt = 2\pi \int_{0}^{\sqrt{3}} \left(t + \frac{2t^{3}}{3} - \frac{t^{5}}{3}\right),dt$$

$$S = 2\pi \left[\frac{t^{2}}{2} + \frac{t^{4}}{6} - \frac{t^{6}}{18}\right]_{0}^{\sqrt{3}} = 2\pi\left(\frac{3}{2} + \frac{3}{2} - \frac{3}{2}\right)$$

$$S = 3\pi$$

Final answer: $3\pi \approx 9.4248$ square units.

Example 6

Rotate $y = \cos x$, $0 \leq x \leq \frac{\pi}{2}$, about the y-axis and find the surface area.

The radius is $x$ and $\frac{dy}{dx} = -\sin x$:

$$S = 2\pi \int_{0}^{\pi/2} x\sqrt{1 + \sin^{2}x},dx$$

This integral is non-elementary: no combination of polynomials, roots, exponentials, logarithms and trig functions has this derivative, because $\sqrt{1 + \sin^{2}x}$ already leads to an elliptic integral. So the honest answer is a number, found by numerical integration:

$$\text{Simpson's rule, } n = 4: \quad S \approx 10.0693$$

$$\text{Simpson's rule, } n = 8: \quad S \approx 10.0742$$

$$\text{High-precision quadrature:} \quad S \approx 10.0745$$

The estimates settle as $n$ grows, which is your evidence that the value is right.

Final answer: $S \approx 10.0745$ square units (no closed form exists).

What If The Integral Has No Closed Form?

Many surface integrals of revolution have no elementary antiderivative, because the square root in $ds$ rarely simplifies. That is normal, and the right response is a careful numeric value, not an invented formula.

A useful contrast: rotate $y = \sin x$, $0 \leq x \leq \pi$, about the x-axis instead. The substitution $u = \cos x$ turns the integral into $2\pi\int_{-1}^{1}\sqrt{1 + u^{2}},du$, which has the exact value $2\pi\left(\sqrt{2} + \ln(1 + \sqrt{2})\right) \approx 14.4236$. The same arch rotated about a different axis (Example 6) has no closed form at all.

Closed forms deserve a numeric check too. For $y = \frac{1}{x}$ on $[1, 2]$ about the x-axis, an expression that circulates online, $2\pi\left(\frac{\sqrt{5}}{2} - \sqrt{2}\right)$, evaluates to about $-1.8610$: a negative area, so it cannot be right. The correct value is

$$S = \pi\left(\sqrt{2} - \frac{\sqrt{17}}{4} + \sinh^{-1}4 - \ln\left(1 + \sqrt{2}\right)\right) \approx 5.0164$$

A ten-second check with a calculator's integration key would have caught the error.

Can A Finite Volume Have An Infinite Surface Area?

Yes. Rotating $y = \frac{1}{x}$ for $x \geq 1$ about the x-axis gives Gabriel's horn, a solid with finite volume and infinite surface area. Forum threads still debate "the paradox of Gabriel's horn," and the calculus settles it with two improper integrals.

$$V = \pi \int_{1}^{\infty} \frac{1}{x^{2}},dx = \lim_{A \to \infty} \pi\left(1 - \frac{1}{A}\right) = \pi$$

$$S = 2\pi \int_{1}^{A} \frac{1}{x}\sqrt{1 + \frac{1}{x^{4}}},dx \geq 2\pi \int_{1}^{A} \frac{1}{x},dx = 2\pi \ln A$$

Since $\ln A \to \infty$, the surface area is infinite while the volume stays at $\pi$. The horn could be filled with $\pi$ cubic units of paint, yet no finite amount could coat its inside with a layer of fixed thickness. Evangelista Torricelli published the finite-volume result in 1644, and it astonished mathematicians of his time. The full story, with the history, is on the Gabriel's horn page.

Why Does Surface Area Of Revolution Matter?

"Curved surfaces cannot be unrolled flat" is the real problem this formula solves. A cylinder or cone can be cut and laid flat, so its area comes from a rectangle or a sector. A sphere, a bowl or a nose cone cannot be flattened without stretching or tearing, so no net and no ruler gives its area.

• The ancient answer was case by case. Archimedes showed in On the Sphere and Cylinder that the surface of a sphere is four times its great circle, with an ingenious argument that worked for the sphere alone.

• The calculus answer is general. $S = 2\pi\int(\text{radius}),ds$ handles any smooth curve and any axis with one method.

• A shortcut exists for symmetric shapes. Pappus's centroid theorem says the area equals the curve's length times the distance its centroid travels, which links this topic to moments. A circle of radius $a$ whose centre sits $R$ from the axis sweeps a torus of area $(2\pi a)(2\pi R) = 4\pi^{2}Ra$.

The same idea grows up in multivariable calculus as the surface integral, and in the catenoid, the soap-film shape that is one of only two minimal surfaces of revolution.

Where Is Surface Area Of Revolution Used In The Real World?

Any object made by spinning a profile has an area worth knowing, because area sets how much material covers it and how fast heat crosses it.

• Tanks, domes and silos: coating, insulation and sheet-metal estimates depend on curved area, not floor area.

• Parabolic reflectors: satellite dishes and headlight mirrors are paraboloids, the surface from Example 3, and their area fixes the reflective material needed.

• Lathe-turned and spun parts: nose cones, bottles, funnels and lampshades are all surfaces of revolution, so their profiles feed straight into this formula.

• Heat exchange: heat loss and cooling scale with surface area, so a designer comparing two vessel shapes compares their surface integrals.

• Torus parts: O-rings, tyres and pipe bends use the Pappus shortcut $4\pi^{2}Ra$ for quick area checks.

Where Do Students Trip Up On Surface Area Of Revolution?

Mistake 1: Using Dx Instead Of Ds

Where it slips in: Setting up the integral quickly, by analogy with the disk method.

Don't do this: Write $S = 2\pi\int y,dx$; for a sphere of radius 2 it gives $4\pi^{2}$ instead of $16\pi$.

The correct way: Always multiply by the full arc-length element $\sqrt{1 + \left(\frac{dy}{dx}\right)^{2}},dx$. The rusher who copies the disk setup gets every curved example wrong except flat lines parallel to the axis.

Mistake 2: Using The Wrong Radius

Where it slips in: Rotating about the y-axis or about a line such as $y = -1$.

Don't do this: Keep $2\pi y$ because that is the version printed first on the formula sheet.

The correct way: Use the distance from the point to the axis: $x$ for the y-axis, $y + 1$ for the line $y = -1$. The first instinct on a y-axis problem is to keep the $y$, and the memorizer who learned one formula by heart is most exposed here; sketching the circle one point travels fixes it every time.

Mistake 3: Adding Or Dropping The End Caps

Where it slips in: Problems that say "surface area of the solid" instead of "area of the surface".

Don't do this: Add flat disks to a surface of revolution answer by default, or leave them off when the problem asks for a closed solid.

The correct way: The integral gives the lateral surface only. Add $\pi r^{2}$ for each flat end only when the question wants the total surface of a closed solid.

Mistake 4: Treating Area Like Volume When Scaling

Where it slips in: Comparison questions such as "the bowl is twice as wide; how much more coating?"

Don't do this: Scale the area by the cube of the size factor, as you would a volume.

The correct way: Areas scale by the square of the factor, volumes by the cube. Double every length of a surface of revolution and its area grows by $4$, not $8$.

Mistake 4 has a famous real-world version. The square-cube law, first described by Galileo in Two New Sciences (1638), says surface area grows with the square of size and volume with the cube. James Watt saw it in practice: scale models of steam engines lost too much heat, because a small engine has far more surface per unit of volume than a full-size one.

Practice Problems On Surface Area Of Revolution

1. Rotate $y = 2x$, $0 \leq x \leq 3$, about the x-axis.
   Answer to Problem 1: $18\sqrt{5},\pi \approx 126.4467$ square units.

2. Rotate $y = x^{3}$, $0 \leq x \leq 1$, about the x-axis.
   Answer to Problem 2: $\frac{\pi}{27}\left(10\sqrt{10} - 1\right) \approx 3.5631$ square units.

3. Rotate $y = \sqrt{4 - x^{2}}$, $-1 \leq x \leq 1$, about the x-axis (a band of a sphere).
   Answer to Problem 3: $8\pi$ square units.

4. Rotate the curve $x = 3\cos t$, $y = 3\sin t$, $0 \leq t \leq \pi$, about the x-axis.
   Answer to Problem 4: $36\pi$ square units.

5. Rotate $y = x^{2}$, $0 \leq x \leq \sqrt{2}$, about the y-axis.
   Answer to Problem 5: $\frac{13\pi}{3} \approx 13.6136$ square units.

6. Compute the surface area of revolution about the x-axis for $y = x^{2}$ on $[0, 4]$.
   Answer to Problem 6: $\frac{\pi}{32}\left(1032\sqrt{65} - \sinh^{-1}8\right) \approx 816.5661$ square units.

The Short Version

• Surface area of revolution is circumference times arc length, added along the curve: $S = 2\pi\int(\text{radius}),ds$.

• The formula comes from frustum bands, which is why it needs $ds$ and not $dx$.

• The radius is the distance from the point to the axis, so it changes when the axis changes.

• Parametric and polar curves use the same idea with their own form of $ds$.

• When the integral has no closed form, give a verified numeric value instead of forcing a formula.

Your Next Move

Work through Practice Problems 1 to 4 first, then try 5 and 6; if the radius trips you up, return to "What Is The Formula For Revolution About The Y-Axis?", and if the square root will not simplify, reread Example 3. Want a live Bhanzu trainer to walk through more surface area problems with you? Book a free demo class:

Read More

• The disk method for volumes

• The washer method for hollow solids

• The cylindrical shell method

• Applications of integration

• Area under a curve