Mathwords logoMathwords

Surface Area of a Surface of Revolution

Surface Area of a Surface of Revolution

The formulas below give the surface area of a surface of revolution. The axis of rotation must be either the x-axis or the y-axis. The curve being rotated can be defined using rectangular, polar, or parametric equations.

 

Formulas for surface area of revolution: rotation about x-axis uses 2πy ds, y-axis uses 2πx ds, with ds in rectangular,...

 

See also

Surface of revolution, arc length

Key Formula

S=2π∫abr(t) (dxdt)2+(dydt)2  dtS = 2\pi \int_a^b r(t)\,\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;dt For y=f(x)y = f(x) rotated about the x-axis: S=2π∫abf(x) 1+[f′(x)]2  dxS = 2\pi \int_a^b f(x)\,\sqrt{1 + \left[f'(x)\right]^2}\;dx For a polar curve r=f(θ)r = f(\theta) rotated about the polar axis: S=2π∫αβf(θ)sin⁡θ [f(θ)]2+[f′(θ)]2  dθS = 2\pi \int_\alpha^\beta f(\theta)\sin\theta\,\sqrt{\left[f(\theta)\right]^2 + \left[f'(\theta)\right]^2}\;d\theta
Where:
  • SS = Surface area of the surface of revolution
  • r(t)r(t) = Distance from the curve to the axis of rotation (the radius of revolution)
  • f(x)f(x) = The function defining the curve in rectangular form
  • f′(x)f'(x) = The derivative of the function with respect to x
  • [a,b][a, b] = The interval of integration (bounds on x, t, or θ)
  • dx/dt,  dy/dtdx/dt, \; dy/dt = Derivatives of the parametric equations with respect to the parameter t
  • f(θ)f(\theta) = The polar function defining the curve

Worked Example

Problem: Find the surface area generated by revolving the curve y = x³ from x = 0 to x = 1 about the x-axis.
Step 1: Identify the formula. Since y = f(x) is rotated about the x-axis, use the rectangular form:
S=2π∫01f(x) 1+[f′(x)]2  dxS = 2\pi \int_0^1 f(x)\,\sqrt{1 + [f'(x)]^2}\;dx
Step 2: Compute the derivative of f(x) = x³.
f′(x)=3x2f'(x) = 3x^2
Step 3: Substitute f(x) and f'(x) into the formula.
S=2π∫01x3 1+9x4  dxS = 2\pi \int_0^1 x^3\,\sqrt{1 + 9x^4}\;dx
Step 4: Use the substitution u = 1 + 9x⁴, so du = 36x³ dx, which gives x³ dx = du/36. When x = 0, u = 1; when x = 1, u = 10.
S=2π∫110u  du36=π18∫110u1/2  duS = 2\pi \int_1^{10} \sqrt{u}\;\frac{du}{36} = \frac{\pi}{18}\int_1^{10} u^{1/2}\;du
Step 5: Evaluate the integral.
S=π18⋅23 u3/2∣110=π27(103/2−1)=π27(1010−1)S = \frac{\pi}{18}\cdot\frac{2}{3}\,u^{3/2}\Big|_1^{10} = \frac{\pi}{27}\left(10^{3/2} - 1\right) = \frac{\pi}{27}\left(10\sqrt{10} - 1\right)
Answer: The surface area is π(1010−1)27≈3.563\dfrac{\pi(10\sqrt{10} - 1)}{27} \approx 3.563 square units.

Another Example

This example uses parametric equations instead of a rectangular function, and it verifies the integral formula against the well-known sphere surface area formula.

Problem: Find the surface area generated by revolving the circle given by the parametric equations x = 3 cos t, y = 3 sin t, for 0 ≤ t ≤ π, about the x-axis. (This produces a sphere of radius 3.)
Step 1: Identify the parametric surface area formula with rotation about the x-axis. Here r(t) = y(t) = 3 sin t (the distance from the curve to the x-axis).
S=2π∫0πy(t) (dxdt)2+(dydt)2  dtS = 2\pi \int_0^{\pi} y(t)\,\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;dt
Step 2: Compute the derivatives: dx/dt = −3 sin t and dy/dt = 3 cos t.
(−3sin⁡t)2+(3cos⁡t)2=9sin⁡2t+9cos⁡2t=3\sqrt{(-3\sin t)^2 + (3\cos t)^2} = \sqrt{9\sin^2 t + 9\cos^2 t} = 3
Step 3: Substitute into the integral.
S=2π∫0π(3sin⁡t)(3)  dt=18π∫0πsin⁡t  dtS = 2\pi \int_0^{\pi} (3\sin t)(3)\;dt = 18\pi \int_0^{\pi} \sin t\;dt
Step 4: Evaluate the integral.
S=18π[−cos⁡t]0π=18π[−(−1)+1]=18π(2)=36πS = 18\pi\left[-\cos t\right]_0^{\pi} = 18\pi\left[-(-1) + 1\right] = 18\pi(2) = 36\pi
Answer: The surface area is 36π36\pi square units, which matches the known formula for a sphere: S=4πr2=4π(9)=36πS = 4\pi r^2 = 4\pi(9) = 36\pi.

Frequently Asked Questions

Why is 2π in the surface area of revolution formula?
The factor 2π comes from the circumference of the circle traced by each point on the curve as it revolves around the axis. Each small arc segment at distance r from the axis sweeps out a band whose width is the arc length element ds and whose 'length' around the axis is 2πr. Multiplying these and integrating gives the total surface area.
What is the difference between surface area of revolution and volume of revolution?
Surface area of revolution measures the outer area of the solid formed by rotating a curve, using the integral of 2πr·ds (where ds is arc length). Volume of revolution measures the space enclosed inside the solid, typically using the disk method (∫ πr² dx) or the shell method (∫ 2πrh dx). The key distinction is that surface area involves the arc length element, while volume does not.
How do you set up the surface area integral when rotating about the y-axis instead of the x-axis?
When rotating about the y-axis, the radius of revolution becomes the horizontal distance from the curve to the y-axis, which is x (or x(t) in parametric form). For y = f(x), the formula becomes S = 2π ∫ x √(1 + [f'(x)]²) dx. Everything else stays the same — you still need the arc length element under the square root.

Surface Area of Revolution vs. Arc Length of a Curve

Surface Area of RevolutionArc Length of a Curve
What it measuresArea of the surface formed by rotating a curve about an axisLength of a curve between two points
Core formula (rectangular)S=2π∫f(x)1+[f′(x)]2 dxS = 2\pi \int f(x)\sqrt{1+[f'(x)]^2}\,dxL=∫1+[f′(x)]2 dxL = \int \sqrt{1+[f'(x)]^2}\,dx
Key differenceMultiplies the arc length element ds by the circumference factor 2πrIntegrates the arc length element ds alone
UnitsSquare units (area)Linear units (length)

Why It Matters

This topic appears in second-semester calculus (Calculus II) and is a standard application of integration. Engineers use surface area of revolution to calculate material needed for rotationally symmetric objects like tanks, nozzles, domes, and pipes. It also reinforces understanding of arc length and parametric integration, both of which are essential for multivariable calculus.

Common Mistakes

Mistake: Forgetting the arc length element and using just dx instead of √(1 + [f'(x)]²) dx.
Correction: Surface area requires integrating along the actual arc of the curve, not just along the x-axis. Always include the full arc length factor ds = √(1 + [f'(x)]²) dx under the integral. Without it, you are not accounting for how steeply the curve rises.
Mistake: Using the wrong radius of revolution — for example, using f(x) when rotating about the y-axis.
Correction: The radius r in the factor 2πr must be the perpendicular distance from the curve to the axis of rotation. When rotating about the x-axis, r = y = f(x). When rotating about the y-axis, r = x. Mixing these up gives an entirely different (and incorrect) surface.

Related Terms