Surface Area of a Solid of Revolution | Free Calculus 2 Lesson (4 College Credits) | UPI Study

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Surface Area of a Solid of Revolution

Calculus 2 · Chapter 4: Applications of IntegrationSection 4.5 · lesson 20 of 549 min · free

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In this lesson

From arc length to the area of a spun surface.

  • See why the building block is a frustum band, not a flat ring
  • Attach the factor 2\pi r to the arc length element ds
  • Set up surfaces about the x-axis and the y-axis
  • Choose the right radius and the right form of ds

Transcript

Everything said in the video, section by section (1,404 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 4.5

Surface Area of Revolution

How much aluminum foil would it take to wrap the outside of a parabolic satellite dish? How much chrome plating coats a turned table leg? How much heat-shield material covers the curved nose cone of a rocket? Each of those questions asks for the surface area of a solid of revolution. Not the volume it encloses. The area of the skin wrapped around it. Section four point two found such volumes by stacking disks. Here the goal is the outer surface itself. And that changes the building block. It's no longer a flat disk. It's a curved band.

The Question

Not the Volume - the Skin

The key realization is this. A surface of revolution is generated by a curve. Not a region. That distinction matters. In the volume sections, you revolved a two-dimensional region and got a solid. Here you revolve a one-dimensional curve, and you get a surface. A shell with no thickness. As the curve revolves, each tiny piece of it traces out a thin ring-shaped band. And the area of that band depends on exactly two things. How far the piece is from the axis. And how long the piece is. The first is just the radius. The second is the arc-length element d s, which you developed in section four point four. So this section is really just a combination. Take d s. Attach a circumference. Integrate.

The Building Block

The Frustum Band

Let's be precise about the band. When a short, nearly straight arc element is revolved about an axis, it does not sweep out a flat ring. It sweeps out a slightly slanted band. The lateral surface of a frustum — which is a cone with its tip sliced off. Look at the right-hand picture. That's the shape one arc element makes when it spins. Now, a standard result from geometry gives the lateral area of such a band. It's two pi, times its average radius, times its slant height. For an infinitesimally short element, the average radius is simply the distance r from the axis to the curve. And the slant height is exactly the arc-length element d s. So the area of one band is two pi r d s. And d s is carried over unchanged from arc length. Root of one plus f prime squared, d x. Or the y-form, root of one plus g prime squared, d y, when the curve is given that way. The surface-area formulas do nothing more than attach the factor two pi r to it.

About the x-Axis

The Radius Is the Height of the Curve

Start with revolution about the x-axis. When the curve y equals f of x is revolved about the x-axis, the distance from the axis to the curve is the height of the curve itself. So the radius r is just y. Which is f of x. Substitute that radius and the d s element into the band area, integrate over the interval, and you have the formula. S equals the integral from a to b, of two pi y, root one plus f prime of x squared, d x. Let's test it on something we can check independently. The line y equals two x, on zero to two, revolved about the x-axis. That forms a cone. Here f prime is two. So one plus f prime squared is one plus four. Five. And the radius is y, which is two x. So S is the integral of two pi times two x, times root five. That's four pi root five, times the integral of x from zero to two. Which is four pi root five, times two. Eight pi root five. About fifty-six point two zero square units. Now here's the satisfying part. We can check that against pure geometry. The lateral area of a cone is pi R L. Where R is the base radius and L is the slant height. Our cone has base radius two times two, which is four. And slant height root of two squared plus four squared. Root twenty. Which is two root five. So pi R L is pi times four times two root five. Eight pi root five. An exact match. The integral is confirmed.

About the y-Axis

Only the Radius Changes

Now revolve about the y-axis instead. And notice how little changes. Revolving the same kind of curve about the y-axis changes only the radius. The distance from the y-axis to the curve is now the horizontal coordinate. So r equals x. The band area becomes two pi x d s. And the integral is set up to match how the curve is described. If the curve is given as x equals g of y, integrate in y. If it's given as y equals f of x, the radius is still x, but the element keeps its x-form. Let's do one. The curve y equals x squared, on zero to one, revolved about the y-axis. The radius from the y-axis is r equals x. With f prime equal to two x, one plus f prime squared is one plus four x squared. So S is the integral from zero to one, of two pi x, root one plus four x squared. And this integral yields to a u-substitution. Let u be one plus four x squared, so d u is eight x d x — meaning x d x is d u over eight. Change the limits: when x is zero, u is one. When x is one, u is five. Work it through, and you get pi over six, times five root five minus one. About five point three three square units. Now notice something important about that setup. The radius switched to x, because the axis switched to the y-axis. But the element d s kept its x-form, because the curve was still given as y equals f of x. Those are two independent choices. Don't let one drag the other along. And one honest note. Not every surface-area integrand cooperates this nicely. Had the curve been y equals x cubed, the integrand would contain x d x against an x-cubed-based substitution — leaving no elementary antiderivative, and forcing a numerical estimate.

Putting It Together

One Template, Two Choices

So let's put it all together, because every problem in this section is one template. S equals the integral of two pi r, d s. That's it. And it's specialized by exactly two choices. First choice: the radius. r is the distance from the axis to the curve. So r equals y for the x-axis. r equals x for the y-axis. And for a shifted axis, y equals k or x equals h, it's the appropriate shifted distance — exactly as in section four point two. Second choice: the element. d s is written in whichever variable matches the curve's description. The x-form for y equals f of x. The y-form for x equals g of y. Getting those two choices right is the entire skill. The rest is just integration. Here's something worth thinking about, and it goes to the heart of why d s matters. Both the volume formula from section four point two, and the surface-area formula here, use the radius r. But volume integrates pi r squared over the flat thickness d x. While surface area integrates two pi r over the slanted d s. Explain why volume uses the flat thickness, but surface area must use the slant. And ask yourself: what would go wrong — and in which direction would the answer err — if someone mistakenly used d x in place of d s in a surface-area calculation?

Key Takeaways

What to Carry Forward

Five things to carry forward. A surface of revolution is generated by a curve, and each arc element sweeps out a frustum band of area two pi r d s. The element d s is taken directly from arc length, section four point four. Surface area simply attaches the factor two pi r to it. About the x-axis, the radius is y. So S is the integral of two pi y, root one plus f prime squared, d x. About the y-axis, the radius is x. So S is the integral of two pi x, root one plus f prime squared, d x — or the y-form when the curve is given as x equals g of y. And setting up the integral correctly is the essential skill. Surface-area integrands often have no elementary antiderivative, and a numerical value is frequently the practical answer once the setup is right. Next up is section four point six — Work. Where the integral accumulates force over distance.

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FAQ

Questions students ask us

About Surface Area of a Solid of Revolution

What is the formula for the surface area of a solid of revolution?
The surface area is S = \int 2\pi r\,ds, where r is the distance from the axis to the curve and ds is the arc length element. About the x-axis, r = y, so S = \int_a^b 2\pi f(x)\sqrt{1+[f'(x)]^2}\,dx. About the y-axis, r = x, so S = \int_a^b 2\pi x\sqrt{1+[f'(x)]^2}\,dx. If the curve is given as x = g(y), use the y-form of ds instead.
Why does surface area of revolution use ds instead of dx?
Surface area uses ds because each revolved arc element sweeps out a slanted frustum band, not a flat ring, and the band's slant height is ds. The lateral area of a frustum is 2\pi times its average radius times its slant height, so each band contributes 2\pi r\,ds. Volume can use the flat thickness dx, but using dx for surface area ignores the slant. Since ds \ge dx, that mistake underestimates the area.
How do you find the surface area of a cone using integration?
Revolve a line through the origin, such as y = 2x on [0,2], about the x-axis and integrate 2\pi y\,ds. Since f'(x) = 2, you get ds = \sqrt{5}\,dx. Then S = \int_0^2 2\pi(2x)\sqrt{5}\,dx = 4\pi\sqrt{5}\cdot 2 = 8\pi\sqrt{5} \approx 56.20 square units. This matches the geometric formula \pi R L with base radius R = 4 and slant height L = 2\sqrt{5}.
How do you set up surface area when revolving y = f(x) about the y-axis?
Use radius r = x but keep the x-form of ds. The radius depends on the axis, while ds depends on how the curve is described. For y = x^2 on [0,1], S = \int_0^1 2\pi x\sqrt{1+4x^2}\,dx. Substituting u = 1+4x^2 gives \frac{\pi}{6}(5\sqrt{5}-1) \approx 5.33 square units. A common mistake is letting one choice drag the other along.

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