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A band's area is its circumference times its slant width, so $S = 2\pi\int y\,ds$ — the arc length element, not $dx$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write the area of a band as circumference times slant width, build the slant element from the derivative and say why a band's width is measured along the surface rather than along the axis, use the distance to the axis of revolution as the radius when that axis is not a coordinate axis, tell the surface integrand apart from the volume integrand symbol by symbol, and explain why every band of equal width on a sphere has equal area.
The last lesson established $ds = \sqrt{1 + (y')^2}\,dx$: the length of a short piece of curve. Integrating it gives the curve's length.
Sweep that same short piece once round an axis and it traces a thin band. The band's area is its length round — the circumference $2\pi y$ — times its width, and its width is the piece itself, $ds$. Everything in this lesson follows from taking that second factor seriously. Every application of the integral is the same move: write what one thin slice contributes, then add the slices up. Get the slice right and the integral writes itself; reach for a remembered formula instead and the first unfamiliar region defeats you.
A band (or zone) is the surface swept by a short piece of the curve: a thin ring, not a flat annulus.
A frustum is a cone with its tip cut off — the exact shape swept by a straight chord. Its lateral area is $\pi(r_1 + r_2)\ell$, with $\ell$ the slant height, and taking $r_1 \approx r_2 = y$ and $\ell = ds$ gives $2\pi y\,ds$. That frustum, not a cylinder, is what makes the derivation honest.
The slant height is the distance measured along the sloping surface. It is the whole difficulty of this lesson: a cylinder's slant height equals its height, and a cone's does not.
The formula. Revolving $y = f(x) \ge 0$ on $[a,b]$ about the $x$-axis gives a surface of area $$S = \int_a^b 2\pi y\,ds = 2\pi\int_a^b f(x)\sqrt{1 + f'(x)^2}\,dx.$$
About the $y$-axis instead, the radius is $x$: $$S = 2\pi\int x\,ds,$$ with $ds$ written in whichever variable is convenient.
Why $ds$ and not $dx$. A band is a piece of the surface, so its width is measured along the surface. The exact shape swept by a straight chord is a frustum, whose lateral area is (average circumference) $\times$ (slant height); in the limit the average circumference is $2\pi y$ and the slant height is $ds$. Replacing $ds$ by $dx$ would be measuring a roof by its floor plan, and it understates every sloping surface.
The two factors do different jobs. $2\pi y$ is how far round; $ds$ is how far along. Revolving about a different line changes only the first; changing the curve changes both.
The integrals are worse than arc length's. They carry the same square root and an extra factor, so elementary answers are rarer still. The sphere is the great exception, and it is an exception for a reason worth knowing: there the two factors cancel.
When the curve dips below the axis, the radius is $|y|$, since a radius is a distance.
Another way: picture
A lampshade, made from a ring of card. Flatten a thin horizontal band of it: you get a long thin strip whose length is the circumference $2\pi y$ and whose width is the slanted distance $ds$ down the shade — not the vertical drop, which is shorter. Cut the band with scissors held horizontally and you cut a longer path than the drop, and that extra is exactly the square root.
Another way: steps
Revolve the semicircle $y = \sqrt{r^2 - x^2}$ about the $x$-axis. Then $$y' = \frac{-x}{\sqrt{r^2 - x^2}}, \qquad \sqrt{1 + (y')^2} = \frac{r}{\sqrt{r^2 - x^2}},$$ and multiplying by the radius $y = \sqrt{r^2 - x^2}$ gives exactly $r$. The integrand is constant.
So $S = 2\pi r\int_{-r}^{r}dx = 4\pi r^2$, and more than that: every band of the same width has the same area, wherever it lies on the sphere. Near the pole the bands are narrow round but very steep, so $ds$ is large; near the equator they are wide round but nearly flat, so $ds$ is small. The two effects cancel exactly, and only for the sphere.
That is Archimedes' theorem on the sphere and cylinder: the lateral area of the sphere equals that of the cylinder that just contains it, $2\pi r \cdot 2r$. He proved it without calculus and asked for the figure on his tombstone. It is also why an equal-area map projection can be made by projecting the globe horizontally onto a cylinder — the cartography rests on the cancellation above.
Using $dx$ instead of $ds$. The single error this lesson exists to prevent. A band's width is along the surface.
Forgetting the radius. $\int 2\pi\,ds$ is a length times a constant, not an area.
Using the wrong radius after changing the axis. About the $y$-axis the radius is $x$; about $y = c$ it is $|y - c|$.
Taking $y$ negative. A radius is a distance. If the curve dips below the axis, use $|y|$ or split the integral.
Confusing this with the volume formula. Volume by plates has $\pi y^2\,dx$: a square and a horizontal thickness. Surface has $2\pi y\,ds$: a first power and a slant width. Every symbol differs.
Expecting the integral to be elementary. It is worse than arc length. Set it up correctly and evaluate numerically when it does not come out.
Side by side, $\pi\int y^2\,dx$ and $2\pi\int y\,ds$ look like variations on a theme, and the natural summary is "volume squares the radius, surface does not". That summary loses the only thing that matters, which is the difference between $dx$ and $ds$.
The reason they differ is geometric, not notational. A plate is a slice cut across the solid: it is a solid object, and its thickness is genuinely horizontal, so $dx$ is right. A band is a strip of the surface: it lies along the surface and follows its slope, so its width is $ds$. One of these things is cut through the solid and the other is peeled off its skin.
A cylinder hides the distinction entirely, because there $ds = dx$, and almost every first example is a cylinder. So the error survives until the first cone, where the formula with $dx$ gives $\pi r h$ instead of $\pi r\ell$ — a plausible number, wrong by the ratio of the slant to the height, with nothing in the arithmetic to signal it.
The check that catches it: a sloping surface is always larger than its shadow. If your answer for a cone does not exceed the area you would get by pretending the surface were vertical, the $ds$ has gone missing.
Revolve $y = r$ (a constant) on $[0, h]$ about the $x$-axis. Then $y' = 0$, so $ds = dx$ exactly.
The one case where $ds$ and $dx$ agree.
$S = 2\pi\displaystyle\int_0^h r\,dx = 2\pi r h$: the familiar lateral area of a cylinder.
Constant integrand.
This is why the error of writing $dx$ for $ds$ survives so long. It is invisible on a cylinder, and every first example is a cylinder.
Revolve $y = \sqrt{x}$ on $[0, 4]$ about the $x$-axis: $y' = \dfrac{1}{2\sqrt x}$, so $ds = \sqrt{1 + \dfrac{1}{4x}}\,dx$.
Build the element from the derivative.
$2\pi y\,ds = 2\pi\sqrt x\sqrt{1 + \dfrac{1}{4x}}\,dx = 2\pi\sqrt{x + \tfrac14}\,dx$ — the radius pulls inside the root and tidies it, which does not usually happen.
A cancellation, though a smaller one than the sphere's.
$S = 2\pi\left[\tfrac23\left(x + \tfrac14\right)^{3/2}\right]_0^4 = \dfrac{\pi}{6}\left(17^{3/2} - 1\right)$.
$y' = \dfrac{1}{\sqrt x}$, so $\sqrt{1 + (y')^{2}} = \sqrt{1 + \dfrac{1}{x}} = \dfrac{\sqrt{x + 1}}{\sqrt x}$.
Derivative, squared, plus one, rooted.
$2\pi y\,ds = 2\pi \cdot 2\sqrt x \cdot \dfrac{\sqrt{x+1}}{\sqrt x}dx = 4\pi\sqrt{x + 1}\,dx$ — the $\sqrt x$ cancels.
Multiply the radius in before integrating.
$S = 4\pi\left[\tfrac23(x+1)^{3/2}\right]_0^3 = \dfrac{8\pi}{3}(8 - 1) = \dfrac{56\pi}{3}$.
The line $y = \dfrac{8}{15}x$ from $x = 0$ to $x = 30$ is revolved about the $x$-axis. The surface area is $k\pi$. Find $k$.
Answer:
Why is the surface area $2\pi\displaystyle\int y\sqrt{1 + (y')^{2}}\,dx$ rather than $2\pi\displaystyle\int y\,dx$?
Match each quantity to the contribution of one slice.
| $\sqrt{1 + (y')^{2}}\,dx$ | $2\pi y\sqrt{1 + (y')^{2}}\,dx$ | $\pi y^{2}\,dx$ | $2\pi x y\,dx$ | |
|---|---|---|---|---|
| The length of the curve $y = f(x)$ | ||||
| The area of the surface it sweeps about the $x$-axis | ||||
| The volume it encloses about the $x$-axis, by plates | ||||
| The volume it sweeps about the $y$-axis, by tubes |
For $y = \dfrac{3}{4}x$ revolved about the $x$-axis, fill in the band's factors at $x = 12$.
| Value | |
|---|---|
| The radius of the band | |
| The slant element per unit of $x$ | |
| Their product |
Put the steps of finding the area of the surface swept by $y = 4x$ about the $x$-axis into the order you do them.
Number the steps in order (write the number in the box):
Revolving $y = \sqrt{16 - x^{2}}$ on $[-4, 4]$ about the $x$-axis gives a sphere. Its surface area is $k\pi$. Find $k$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The line $y = \dfrac{5}{12}x$ from $x = 0$ to $x = 48$ is revolved about the $x$-axis. The surface area is $k\pi$. Find $k$.
Answer:
You can set up the area of any surface of revolution and say why the square root is there. Say in your own words why the error of writing $dx$ for $ds$ is invisible on a cylinder. Next: work, force and average value — the applications that are not about shape at all.
10. Your turn: the area swept by $y = 2\sqrt{x}$ on $[0, 3]$ about the $x$-axis, step 3