Areas Of Surfaces Of Revolution

When an arc is revolved about a line, it sweeps out a surface of revolution. Every thin piece of arc $d s$ traces a narrow collar whose area is its circumference times its width, $d S = 2 π ρ d s$ , where $ρ$ is the distance from $d s$ to the axis. Adding the collars turns a surface-area problem into a single integral.

Item	Statement
Surface element	$d S = 2 π ρ d s$ , with $ρ$ the distance from $d s$ to the axis
Arc length element	$d s = \sqrt{(d x)^{2} + (d y)^{2}}$
In terms of $x$	$d s = \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$
In terms of $y$	$d s = \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y$
About the $x$ -axis	$ρ = y$ , so $S = \int_{a}^{b} 2 π y \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$
About the $y$ -axis	$ρ = x$ , so $S = \int_{c}^{d} 2 π x \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y$
Cone, lateral	$π r L$ ( $r$ = base radius, $L$ = slant height)
Cone of radius $r$ , height $h$	$π r \sqrt{r^{2} + h^{2}}$
Frustum, lateral	$π (r_{1} + r_{2}) L$
Sphere of radius $R$	$4 π R^{2}$
Zone of a sphere of width $h$	$2 π R h$ , wherever the zone sits

The Surface Area Element

When a curve $y = f (x)$ is revolved about the $x$ -axis, it generates a surface of revolution, as in the figure below. Each point of the curve travels around a circle centred on the axis, so the surface is built out of circles whose radii are the heights $y$ of the curve.

The curve y = f(x) drawn above the x-axis together with the surface of revolution it generates when the curve is turned about the x-axis, each point of the curve tracing a circle centred on the axis. — Revolving the curve $y = f (x)$ about the $x$ -axis generates a surface of revolution.

We can use the arc length formula of Section 8.5 to calculate the area of such a surface.

The arc length element is denoted by $d s$ with a lower case $s$ , and the surface area element is denoted by $d S$ with a capital $S$ .

The surface area of the "collar" generated by rotating an element $d s$ of the curve about the $x$ -axis is $2 π y d s$ , where $y$ is the distance of the midpoint of $d s$ from the $x$ -axis. The collar is a very short band: unroll it and it is a strip of length $2 π y$ (the circumference of the circle it lies on) and width $d s$ .

A short element ds of the curve, a distance y above the x-axis, sweeping out a narrow collar of the surface of revolution whose circumference is 2 pi y. — The collar swept out by the element $d s$ has radius $y$ and width $d s$ , so $d S = 2 π y d s$ .

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{dS=\underbrace{2\pi\overbrace{y}^{\text{radius}}}_{\text{circumference}}ds}\tag{i}

Why the collar has area

2 π y d s

To justify the formula, we need the lateral surface area of the frustum of a right circular cone. First consider a right circular cone of base radius $r$ and slant height $L$ .

If we cut the cone along the highlighted line and flatten it, we get a circular sector of radius $L$ and some central angle $θ$ . The arc of the sector was the base circle of the cone, so its length is $2 π r$ .

The area of a circular sector of radius $L$ and central angle $θ$ (in radians) is

S = \frac{1}{2} L^{2} θ .

Because $L θ = 2 π r$ , we can write

S = \frac{1}{2} L^{2} \cdot \frac{2 π r}{L} = π r L .

Therefore the lateral surface area of a cone with base radius $r$ and slant height $L$ is $π r L$ .

Now we can find the lateral surface area of the frustum of a cone. Extending the frustum's slanted side upward completes it to a full cone with apex $O$ . Let $r_{1}$ and $r_{2}$ be the radii of the top and bottom rims of the frustum ( $r_{1} < r_{2}$ ), let $l$ be the slant height from the apex to the top rim, and let $L$ be the slant height of the frustum itself, so that $l + L$ is the slant height from the apex to the bottom rim.

The lateral surface of the frustum is the lateral surface of the large cone (radius $r_{2}$ , slant height $l + L$ ) minus the lateral surface of the small cone removed from its top (radius $r_{1}$ , slant height $l$ ):

S = π r_{2} (l + L) - π r_{1} l .

By similar triangles, $\frac{l}{r_{1}} = \frac{l + L}{r_{2}}$ , which gives $l r_{2} = r_{1} l + r_{1} L$ , that is, $r_{1} L = l (r_{2} - r_{1})$ . Substituting,

S = π r_{2} l + π r_{2} L - π r_{1} l = π [l (r_{2} - r_{1})] + π r_{2} L = π r_{1} L + π r_{2} L,

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{S=\pi(r_{1}+r_{2})L.}

Now in the collar above, the two rim radii $r_{1}$ and $r_{2}$ are nearly equal, both approaching $y$ , and the slant height is $L = d s$ . Thus

d S = π (y + y) d s = 2 π y d s .

Revolving About the x-Axis

Adding up the collars between the planes $x = a$ and $x = b$ gives the total surface area

S = \int_{a}^{b} d S = \int_{a}^{b} 2 π y d s .

Because $d s = \sqrt{(d x)^{2} + (d y)^{2}} = \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ , the total surface is

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{S=\int_{a}^{b}2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx.}

Since $y = f (x)$ and dy/dx=f'(x), the formula can also be written as

S=\int_{a}^{b}2\pi f(x)\sqrt{1+\left[f'(x)\right]^{2}}\,dx.

Revolving About the y-Axis

Analogously, if the surface of revolution is generated by revolving a curve about the $y$ -axis, the radius of the collar is the horizontal distance $x$ , so

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{dS=2\pi x\,ds}\tag{ii}

If the curve $x = g (y)$ , $c \leq y \leq d$ , is revolved about the $y$ -axis, then since $d s = \sqrt{(d x)^{2} + (d y)^{2}} = \sqrt{{(\frac{d x}{d y})}^{2} + 1} d y$ , the surface area is

S=\int_{c}^{d}2\pi x\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}\,dy=\int_{c}^{d}2\pi g(y)\sqrt{1+\left[g'(y)\right]^{2}}\,dy.

A curve revolved about the y-axis, with the element ds lying a horizontal distance x from the axis and sweeping out a collar of radius x. — If the surface is generated by revolving about the $y$ -axis, then $d S = 2 π x d s$ .

We may combine (i) and (ii) into a single statement:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{dS=2\pi\rho\,ds,}

where $ρ$ is the distance of $d s$ from the axis of rotation. The only decisions you have to make are which variable to integrate in and what $ρ$ is.

Worked Examples: Revolving About the x-Axis

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

The curve y = square root of x from x = 0 to x = 2 and the bowl-shaped surface it generates when revolved about the x-axis. — The surface formed by rotating $y = \sqrt{x}$ , $0 \leq x \leq 2$ , about the $x$ -axis.

Solution

To find the surface area, we evaluate

S=\int_{a}^{b}2\pi y\sqrt{1+\left(y'\right)^{2}}\:dx

with

a=0,\quad b=2,\quad y=\sqrt{x},\quad\text{and}\quad y'=\frac{1}{2\sqrt{x}}.

\begin{aligned} S &= \int_{0}^{2}2\pi\sqrt{x}\sqrt{1+\frac{1}{4x}}\:dx\\ &= 2\pi\int_{0}^{2}\sqrt{x}\sqrt{\frac{4x+1}{4x}}\:dx\\ &= 2\pi\int_{0}^{2}\frac{\sqrt{x}\,\sqrt{4x+1}}{2\sqrt{x}}\:dx\\ &= \pi\int_{0}^{2}\sqrt{4x+1}\,dx. \end{aligned}

A second, smaller view of the surface generated by revolving y = square root of x on the interval from 0 to 2 about the x-axis, accompanying the evaluation of the integral. — A closer view of the surface whose area is being computed.

To find the last integral, let $u = 4 x + 1$ . Then $d u = 4 d x$ .

The lower limit of integration: when $x = 0$ , $u = 1$ .

The upper limit of integration: when $x = 2$ , $u = 9$ .

Therefore

\begin{aligned} S &= \pi\int_{1}^{9}\sqrt{u}\ \overbrace{\frac{1}{4}\,du}^{dx}\\ &= \frac{\pi}{4}\cdot\left.\frac{2}{3}u^{3/2}\right|_{1}^{9}\\ &= \frac{\pi}{6}\left(9^{3/2}-1^{3/2}\right)\\ &= \frac{\pi}{6}\left(27-1\right)\\ &= \frac{26\pi}{6}=\frac{13\pi}{3}. \end{aligned}

Find the surface area of a sphere of radius $R$ .

Solution

The surface of this sphere can be generated by revolving the semicircle $y = \sqrt{R^{2} - x^{2}}$ about the $x$ -axis.

Since

\frac{d y}{d x} = \frac{- 2 x}{2 \sqrt{R^{2} - x^{2}}} = - \frac{x}{\sqrt{R^{2} - x^{2}}},

and the distance of $d s$ from the axis of rotation (the $x$ -axis) is $y$ , we have

\begin{aligned} dS &= 2\pi y\sqrt{(dx)^{2}+(dy)^{2}}\\ &= 2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx\\ &= 2\pi\underbrace{\sqrt{R^{2}-x^{2}}}_{y}\sqrt{1+\left(-\frac{x}{\sqrt{R^{2}-x^{2}}}\right)^{2}}\,dx\\ &= 2\pi\sqrt{R^{2}-x^{2}}\sqrt{\frac{R^{2}}{R^{2}-x^{2}}}\,dx\\ &= 2\pi R\,dx. \end{aligned}

The height and the stretching factor cancel exactly, leaving a constant. Because $x$ varies between $- R$ and $R$ , the surface area is

S = \int_{- R}^{R} d S = \int_{- R}^{R} 2 π R d x = 2 π R x |_{- R}^{R} = 4 π R^{2} .

The same result is obtained if we revolve the semicircle $x = \sqrt{R^{2} - y^{2}}$ about the $y$ -axis.

In this case

d s = \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y,

the distance of $d s$ from the $y$ -axis is $x$ , and $y$ varies between $- R$ and $R$ . That is,

\begin{aligned} S &= \int_{-R}^{R}2\pi\underbrace{\sqrt{R^{2}-y^{2}}}_{x}\sqrt{1+\frac{y^{2}}{R^{2}-y^{2}}}\ dy\\ &= \int_{-R}^{R}2\pi R\,dy=4\pi R^{2}. \end{aligned}

Find the lateral surface area of a cone with base radius $r$ and height $h$ . (The lateral surface excludes the base area.)

Solution

Consider the line $y = \frac{r}{h} x$ , $0 \leq x \leq h$ . If this line segment is revolved about the $x$ -axis, it generates a cone whose base radius is $r$ and whose height is $h$ .

By the surface area formula,

\begin{aligned} S &= \int_{0}^{h}2\pi\left(\overbrace{\frac{r}{h}x}^{y}\right)\sqrt{1+\left(\underbrace{\frac{r}{h}}_{y'}\right)^{2}}\,dx\\ &= \frac{2\pi r}{h}\int_{0}^{h}\sqrt{\frac{r^{2}+h^{2}}{h^{2}}}\ x\,dx\\ &= \frac{2\pi r}{h^{2}}\sqrt{r^{2}+h^{2}}\left[\frac{x^{2}}{2}\right]_{0}^{h}\\ &= \pi r\sqrt{r^{2}+h^{2}}. \end{aligned}

Notice that $\sqrt{r^{2} + h^{2}}$ is the slant height of the cone. This means that

lateral surface area of a cone = π \times (base radius) \times (slant height),

which agrees with the formula $π r L$ derived at the start of this section.

Worked Examples: Revolving About the y-Axis

Find the area of the surface generated by revolving the curve $y = x^{2}$ ( $1 \leq x \leq 2$ ) about the $y$ -axis.

Solution

The distance from $d s$ to the $y$ -axis is $x$ . So $d S = 2 π x d s$ .

Method 1. We can use $x$ as the variable of integration. Then

\begin{aligned} dS &= 2\pi x\,ds\\ &= 2\pi x\sqrt{1+(y')^{2}}\,dx\\ &= 2\pi x\sqrt{1+(2x)^{2}}\,dx. \end{aligned} TikZ figure

and

S = \int_{1}^{2} d S = \int_{1}^{2} 2 π x \sqrt{1 + 4 x^{2}} d x .

Now let $u = 1 + 4 x^{2}$ . Then $d u = 8 x d x$ .

The lower limit of integration: when $x = 1$ , $u = 5$ .

The upper limit of integration: when $x = 2$ , $u = 17$ .

Therefore,

\begin{aligned} S &= \int_{5}^{17}2\pi\sqrt{u}\ \overbrace{\frac{1}{8}\,du}^{x\,dx}\\ &= \frac{\pi}{4}\cdot\frac{2}{3}u^{3/2}\bigg|_{5}^{17}\\ &= \frac{\pi}{6}\left(17^{3/2}-5^{3/2}\right)\\ &\approx 30.8465. \end{aligned}

Method 2. We can use $y$ as the variable of integration. Then $x = \sqrt{y}$ , and since $d x / d y = 1 / (2 \sqrt{y})$ we have

\begin{aligned} dS &= 2\pi x\sqrt{(dx)^{2}+(dy)^{2}}\\ &= 2\pi\overbrace{\sqrt{y}}^{x}\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}\,dy\\ &= 2\pi\sqrt{y}\sqrt{1+\frac{1}{4y}}\,dy\\ &= 2\pi\sqrt{y}\ \frac{\sqrt{1+4y}}{2\sqrt{y}}\,dy\\ &= \pi\sqrt{1+4y}\,dy. \end{aligned}

Because $1 \leq x \leq 2$ , the variable $y$ varies between $1$ and $4$ , and thus

S = \int_{1}^{4} π \sqrt{1 + 4 y} d y .

Let $u = 1 + 4 y$ . Then $d u = 4 d y$ .

The lower limit of integration: when $y = 1$ , $u = 5$ .

The upper limit of integration: when $y = 4$ , $u = 17$ .

Therefore,

\begin{aligned} S &= \int_{5}^{17}\pi\sqrt{u}\ \overbrace{\frac{1}{4}\,du}^{dy}\\ &= \frac{\pi}{4}\cdot\frac{2}{3}u^{3/2}\bigg|_{5}^{17}\\ &= \frac{\pi}{6}\left(17^{3/2}-5^{3/2}\right)\\ &\approx 30.8465, \end{aligned}

as before. Both methods lead to the same integral $\frac{π}{6} u^{3 / 2}$ evaluated from $5$ to $17$ , which is a useful check.

Find the area of the surface formed by revolving the loop $9 x^{2} = y (3 - y)^{2}$ about the $y$ -axis.

The loop of the curve 9x squared equals y times (3 minus y) squared, running from the origin up to the point where the curve crosses itself on the y-axis at y = 3. — The loop of $9 x^{2} = y (3 - y)^{2}$ , revolved about the $y$ -axis.

Solution

For the right part of the curve, when $0 \leq y \leq 3$ , we have

x = \frac{1}{3} (3 - y) \sqrt{y} .

Since

\frac{d x}{d y} = - \frac{1}{3} \sqrt{y} + \frac{3 - y}{6 \sqrt{y}} = \frac{- 2 y + (3 - y)}{6 \sqrt{y}} = \frac{3 (1 - y)}{6 \sqrt{y}} = \frac{1 - y}{2 \sqrt{y}},

the differential of arc length is

\begin{aligned} ds &= \sqrt{(dx)^{2}+(dy)^{2}}\\ &= \sqrt{\left(\frac{dx}{dy}\right)^{2}+1}\ dy\\ &= \sqrt{\frac{(1-y)^{2}}{4y}+1}\ dy\\ &= \sqrt{\frac{(1-2y+y^{2})+4y}{4y}}\ dy\\ &= \sqrt{\frac{1+2y+y^{2}}{4y}}\ dy\\ &= \sqrt{\frac{(1+y)^{2}}{4y}}\ dy\\ &= \frac{1+y}{2\sqrt{y}}\,dy, \end{aligned}

and

\begin{aligned} dS &= 2\pi\overbrace{x}^{\text{radius}}\ ds\\ &= 2\pi\cdot\overbrace{\frac{1}{3}(3-y)\sqrt{y}}^{x}\cdot\overbrace{\frac{1+y}{2\sqrt{y}}\,dy}^{ds}\\ &= \frac{\pi}{3}(3-y)(1+y)\,dy\\ &= \frac{\pi}{3}(3+2y-y^{2})\,dy. \end{aligned}

The awkward $\sqrt{y}$ cancels, which is what makes this curve pleasant. Because $y$ varies between $0$ and $3$ , the surface area is

\begin{aligned} S &= \int_{0}^{3}\frac{\pi}{3}(3+2y-y^{2})\,dy\\ &= \frac{\pi}{3}\left[3y+y^{2}-\frac{1}{3}y^{3}\right]_{0}^{3}\\ &= \frac{\pi}{3}\left[9+9-9\right]\\ &= 3\pi. \end{aligned}

Exercises

Find the area of the surface generated by revolving $y = 2 x$ , $0 \leq x \leq 3$ , about the $x$ -axis, and check your answer against the formula $π r L$ for the lateral area of a cone.

Solution

Here $y = 2 x$ and y'=2, so \sqrt{1+(y')^{2}}=\sqrt{5} is constant:

\begin{aligned} S &= \int_{0}^{3}2\pi(2x)\sqrt{5}\,dx\\ &= 4\sqrt{5}\,\pi\left[\frac{x^{2}}{2}\right]_{0}^{3}\\ &= 4\sqrt{5}\,\pi\cdot\frac{9}{2}=18\sqrt{5}\,\pi\approx126.44. \end{aligned}

Check. The solid is a cone of base radius $r = 2 (3) = 6$ and height $h = 3$ , so its slant height is $L = \sqrt{6^{2} + 3^{2}} = \sqrt{45} = 3 \sqrt{5}$ and

π r L = π (6) (3 \sqrt{5}) = 18 \sqrt{5} π,

which agrees.

Find the area of the surface generated by revolving $y = x^{3}$ , $0 \leq x \leq 1$ , about the $x$ -axis.

Solution

Here y'=3x^{2}, so 1+(y')^{2}=1+9x^{4} and

S = \int_{0}^{1} 2 π x^{3} \sqrt{1 + 9 x^{4}} d x .

Let $u = 1 + 9 x^{4}$ . Then $d u = 36 x^{3} d x$ , so $x^{3} d x = \frac{d u}{36}$ . When $x = 0$ , $u = 1$ ; when $x = 1$ , $u = 10$ . Therefore

\begin{aligned} S &= 2\pi\int_{1}^{10}\sqrt{u}\ \frac{du}{36}\\ &= \frac{\pi}{18}\cdot\frac{2}{3}u^{3/2}\bigg|_{1}^{10}\\ &= \frac{\pi}{27}\left(10^{3/2}-1\right)\\ &= \frac{\pi}{27}\left(10\sqrt{10}-1\right)\approx3.5631. \end{aligned}

The arc $y = \sqrt{4 - x^{2}}$ , $- 1 \leq x \leq 1$ , is revolved about the $x$ -axis. Find the area of the band (a zone) that it sweeps out.

Solution

From $y = \sqrt{4 - x^{2}}$ we get y'=\dfrac{-x}{\sqrt{4-x^{2}}}, so

1+(y')^{2}=1+\frac{x^{2}}{4-x^{2}}=\frac{4}{4-x^{2}},\qquad \sqrt{1+(y')^{2}}=\frac{2}{\sqrt{4-x^{2}}}.

Hence the integrand collapses to a constant:

2\pi y\sqrt{1+(y')^{2}}=2\pi\sqrt{4-x^{2}}\cdot\frac{2}{\sqrt{4-x^{2}}}=4\pi,

and

S = \int_{- 1}^{1} 4 π d x = 8 π \approx 25.13 .

This is $2 π R h$ with $R = 2$ and $h = 2$ , the width of the slab.

The curve $x = 2 \sqrt{y}$ , $0 \leq y \leq 3$ , is revolved about the $y$ -axis. Find the area of the resulting surface.

Solution

Here the curve is already given as $x = g (y)$ , so integrate in $y$ . We have $\frac{d x}{d y} = \frac{1}{\sqrt{y}}$ , so

1 + {(\frac{d x}{d y})}^{2} = 1 + \frac{1}{y} = \frac{y + 1}{y} .

The distance from $d s$ to the $y$ -axis is $x = 2 \sqrt{y}$ , so

\begin{aligned} S &= \int_{0}^{3}2\pi\left(2\sqrt{y}\right)\sqrt{\frac{y+1}{y}}\,dy\\ &= \int_{0}^{3}4\pi\sqrt{y}\cdot\frac{\sqrt{y+1}}{\sqrt{y}}\,dy\\ &= 4\pi\int_{0}^{3}\sqrt{y+1}\,dy\\ &= 4\pi\cdot\frac{2}{3}(y+1)^{3/2}\bigg|_{0}^{3}\\ &= \frac{8\pi}{3}\left(4^{3/2}-1^{3/2}\right)=\frac{8\pi}{3}(8-1)=\frac{56\pi}{3}\approx58.64. \end{aligned}

The segment joining $(1, 0)$ to $(3, 4)$ is revolved about the $y$ -axis. Find the area of the frustum it generates, and verify the answer with the formula $S = π (r_{1} + r_{2}) L$ .

Solution

The line through $(1, 0)$ and $(3, 4)$ has slope $\frac{4 - 0}{3 - 1} = 2$ , so $y = 2 (x - 1)$ , that is, $x = 1 + \frac{y}{2}$ for $0 \leq y \leq 4$ . Then $\frac{d x}{d y} = \frac{1}{2}$ and

\sqrt{1 + {(\frac{d x}{d y})}^{2}} = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2} .

Therefore

\begin{aligned} S &= \int_{0}^{4}2\pi\left(1+\frac{y}{2}\right)\frac{\sqrt{5}}{2}\,dy\\ &= \pi\sqrt{5}\int_{0}^{4}\left(1+\frac{y}{2}\right)dy\\ &= \pi\sqrt{5}\left[y+\frac{y^{2}}{4}\right]_{0}^{4}\\ &= \pi\sqrt{5}\,(4+4)=8\sqrt{5}\,\pi\approx56.20. \end{aligned}

Check. The rim radii are $r_{1} = 1$ and $r_{2} = 3$ , and the slant height is $L = \sqrt{(3 - 1)^{2} + (4 - 0)^{2}} = \sqrt{20} = 2 \sqrt{5}$ . Then

π (r_{1} + r_{2}) L = π (1 + 3) (2 \sqrt{5}) = 8 \sqrt{5} π,

as expected.

Find the area of the surface generated by revolving the catenary $y = \cosh x$ , $0 \leq x \leq 1$ , about the $x$ -axis. Give the exact value and a decimal approximation.

Solution

Since y'=\sinh x and $\cosh^{2} x - \sinh^{2} x = 1$ , we get

1+(y')^{2}=1+\sinh^{2}x=\cosh^{2}x,\qquad\sqrt{1+(y')^{2}}=\cosh x,

because $\cosh x > 0$ for all $x$ . Therefore

S = \int_{0}^{1} 2 π \cosh x \cdot \cosh x d x = 2 π \int_{0}^{1} \cosh^{2} x d x .

Using the identity $\cosh^{2} x = \frac{1 + \cosh 2 x}{2}$ ,

\begin{aligned} S &= 2\pi\int_{0}^{1}\frac{1+\cosh 2x}{2}\,dx\\ &= \pi\left[x+\frac{\sinh 2x}{2}\right]_{0}^{1}\\ &= \pi\left(1+\frac{\sinh 2}{2}\right)\\ &= \frac{\pi}{4}\left(4+e^{2}-e^{-2}\right)\approx8.8386. \end{aligned}

The last step uses $\sinh 2 = \frac{e^{2} - e^{- 2}}{2}$ .

Find the area of the surface generated by revolving $y = \sqrt{2 x - x^{2}}$ , $0 \leq x \leq 2$ , about the $x$ -axis. Identify the surface.

Solution

Squaring, $y^{2} = 2 x - x^{2}$ , so $x^{2} - 2 x + y^{2} = 0$ , that is, $(x - 1)^{2} + y^{2} = 1$ . The curve is the upper half of the circle of radius $1$ centred at $(1, 0)$ , so revolving it about the $x$ -axis produces a sphere of radius $1$ . We expect $4 π$ .

Differentiating,

y'=\frac{2-2x}{2\sqrt{2x-x^{2}}}=\frac{1-x}{\sqrt{2x-x^{2}}},

1+(y')^{2}=\frac{(2x-x^{2})+(1-2x+x^{2})}{2x-x^{2}}=\frac{1}{2x-x^{2}}.

Hence the integrand is constant:

2\pi y\sqrt{1+(y')^{2}}=2\pi\sqrt{2x-x^{2}}\cdot\frac{1}{\sqrt{2x-x^{2}}}=2\pi,

and

S = \int_{0}^{2} 2 π d x = 4 π .

The tangent line is vertical at $x = 0$ and $x = 2$ , so \sqrt{1+(y')^{2}} blows up there, but the factor $y$ vanishes at exactly the same rate and the product is the constant $2 π$ . The answer $4 π = 4 π (1)^{2}$ is the surface area of a sphere of radius $1$ , as predicted.

The curve $x = \frac{y^{2}}{4} - \frac{1}{2} \ln y$ , $1 \leq y \leq 2$ , is revolved about the $x$ -axis. Find the area of the surface.

Solution

The curve is given as $x = g (y)$ , so we integrate in $y$ ; but the axis is the $x$ -axis, so the radius is $ρ = y$ . First,

\frac{d x}{d y} = \frac{y}{2} - \frac{1}{2 y} .

Then

\begin{aligned} 1+\left(\frac{dx}{dy}\right)^{2} &= 1+\frac{y^{2}}{4}-\frac{1}{2}+\frac{1}{4y^{2}}\\ &= \frac{y^{2}}{4}+\frac{1}{2}+\frac{1}{4y^{2}}\\ &= \left(\frac{y}{2}+\frac{1}{2y}\right)^{2}. \end{aligned}

Since $\frac{y}{2} + \frac{1}{2 y} > 0$ on $[1, 2]$ , we get $d s = (\frac{y}{2} + \frac{1}{2 y}) d y$ and

\begin{aligned} S &= \int_{1}^{2}2\pi y\left(\frac{y}{2}+\frac{1}{2y}\right)dy\\ &= 2\pi\int_{1}^{2}\left(\frac{y^{2}}{2}+\frac{1}{2}\right)dy\\ &= \pi\int_{1}^{2}\left(y^{2}+1\right)dy\\ &= \pi\left[\frac{y^{3}}{3}+y\right]_{1}^{2}\\ &= \pi\left[\left(\frac{8}{3}+2\right)-\left(\frac{1}{3}+1\right)\right]=\pi\left(\frac{14}{3}-\frac{4}{3}\right)=\frac{10\pi}{3}\approx10.47. \end{aligned}

Find the area of the surface generated by revolving $y = \frac{2}{3} x^{3 / 2}$ , $0 \leq x \leq 3$ , about the $y$ -axis. Use the substitution u=1+(y')^{2}.

Solution

Here y'=x^{1/2}, so (y')^{2}=x and

u=1+(y')^{2}=1+x.

The radius about the $y$ -axis is $x$ , so

S = \int_{0}^{3} 2 π x \sqrt{1 + x} d x .

With $u = 1 + x$ we have $x = u - 1$ and $d x = d u$ ; when $x = 0$ , $u = 1$ , and when $x = 3$ , $u = 4$ . Therefore

\begin{aligned} S &= 2\pi\int_{1}^{4}(u-1)\sqrt{u}\,du\\ &= 2\pi\int_{1}^{4}\left(u^{3/2}-u^{1/2}\right)du\\ &= 2\pi\left[\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}\right]_{1}^{4}\\ &= 2\pi\left[\left(\frac{2}{5}\cdot32-\frac{2}{3}\cdot8\right)-\left(\frac{2}{5}-\frac{2}{3}\right)\right]\\ &= 2\pi\left[\left(\frac{64}{5}-\frac{16}{3}\right)-\left(\frac{2}{5}-\frac{2}{3}\right)\right]\\ &= 2\pi\left[\frac{62}{5}-\frac{14}{3}\right]=2\pi\cdot\frac{186-70}{15}=\frac{232\pi}{15}\approx48.59. \end{aligned}

Let $R > 0$ and let $- R \leq a < b \leq R$ . Show that the zone of the sphere $x^{2} + y^{2} = R^{2}$ (revolved about the $x$ -axis) lying between the planes $x = a$ and $x = b$ has area $2 π R (b - a)$ , no matter where the slab sits. Deduce that this equals the lateral area of the circumscribing cylinder over the same slab.

Solution

The upper half of the sphere is generated by $y = \sqrt{R^{2} - x^{2}}$ . As in the sphere example of this section,

y'=\frac{-x}{\sqrt{R^{2}-x^{2}}},\qquad 1+(y')^{2}=\frac{R^{2}}{R^{2}-x^{2}},

2\pi y\sqrt{1+(y')^{2}}=2\pi\sqrt{R^{2}-x^{2}}\cdot\frac{R}{\sqrt{R^{2}-x^{2}}}=2\pi R.

The integrand is the constant $2 π R$ , independent of $x$ . Hence

S = \int_{a}^{b} 2 π R d x = 2 π R (b - a),

which depends only on the width $h = b - a$ of the slab and not on where the slab is placed. Writing $S = 2 π R h$ makes this clear.

The circumscribing cylinder has radius $R$ and axis the $x$ -axis. The portion of it between $x = a$ and $x = b$ is a cylinder of radius $R$ and length $b - a$ , whose lateral area is (circumference) $\times$ (length) $= 2 π R (b - a)$ . The two areas are equal, which is Archimedes' hat-box theorem. Taking $a = - R$ and $b = R$ recovers $S = 2 π R (2 R) = 4 π R^{2}$ for the whole sphere.

The astroid $x^{2 / 3} + y^{2 / 3} = a^{2 / 3}$ ( $a > 0$ ) is revolved about the $x$ -axis. Find the area of the resulting surface.

Solution

On the right half, $0 \leq x \leq a$ , solve for $y$ :

y = {(a^{2 / 3} - x^{2 / 3})}^{3 / 2} .

Then

y'=\frac{3}{2}\left(a^{2/3}-x^{2/3}\right)^{1/2}\cdot\left(-\frac{2}{3}x^{-1/3}\right)=-\frac{\left(a^{2/3}-x^{2/3}\right)^{1/2}}{x^{1/3}},

1+(y')^{2}=1+\frac{a^{2/3}-x^{2/3}}{x^{2/3}}=\frac{a^{2/3}}{x^{2/3}},\qquad\sqrt{1+(y')^{2}}=\frac{a^{1/3}}{x^{1/3}}.

Therefore the area swept out by the right half is

S_{right} = \int_{0}^{a} 2 π {(a^{2 / 3} - x^{2 / 3})}^{3 / 2} \frac{a^{1 / 3}}{x^{1 / 3}} d x .

Let $w = a^{2 / 3} - x^{2 / 3}$ . Then $d w = - \frac{2}{3} x^{- 1 / 3} d x$ , that is, $\frac{d x}{x^{1 / 3}} = - \frac{3}{2} d w$ . When $x = 0$ , $w = a^{2 / 3}$ ; when $x = a$ , $w = 0$ . Hence

\begin{aligned} S_{\text{right}} &= 2\pi a^{1/3}\int_{a^{2/3}}^{0}w^{3/2}\left(-\frac{3}{2}\right)dw\\ &= 3\pi a^{1/3}\int_{0}^{a^{2/3}}w^{3/2}\,dw\\ &= 3\pi a^{1/3}\cdot\frac{2}{5}\left(a^{2/3}\right)^{5/2}\\ &= \frac{6\pi}{5}a^{1/3}\cdot a^{5/3}=\frac{6\pi}{5}a^{2}. \end{aligned}

The left half ( $- a \leq x \leq 0$ ) contributes the same amount by symmetry, so the total surface area is

S = \frac{12 π}{5} a^{2} .

By the symmetry of the astroid in $x$ and $y$ , revolving it about the $y$ -axis gives the same answer.

The curve $y = \frac{1}{x}$ , $1 \leq x \leq b$ , is revolved about the $x$ -axis, producing a horn. Show that as $b \to \infty$ the volume of the horn tends to the finite number $π$ , while its surface area grows without bound.

Solution

Volume. By the disk method of Section 8.2,

V (b) = \int_{1}^{b} π y^{2} d x = π \int_{1}^{b} \frac{d x}{x^{2}} = π {[- \frac{1}{x}]}_{1}^{b} = π (1 - \frac{1}{b}) .

As $b \to \infty$ , $\frac{1}{b} \to 0$ , so $V (b) \to π$ , a finite number.

Surface area. Here y'=-\dfrac{1}{x^{2}}, so

S (b) = \int_{1}^{b} 2 π \cdot \frac{1}{x} \sqrt{1 + \frac{1}{x^{4}}} d x .

This integral has no elementary closed form worth writing down, but we do not need one. Since $\sqrt{1 + \frac{1}{x^{4}}} \geq 1$ for every $x$ , the integrand satisfies

\frac{2 π}{x} \sqrt{1 + \frac{1}{x^{4}}} \geq \frac{2 π}{x},

and therefore

S (b) \geq \int_{1}^{b} \frac{2 π}{x} d x = 2 π \ln b .

Since $\ln b \to \infty$ as $b \to \infty$ , we conclude $S (b) \to \infty$ .

So the horn encloses a volume approaching $π$ yet has unbounded surface area. The reason the two behave differently is the exponent: the volume integrand carries $y^{2} = 1 / x^{2}$ , whose improper integral converges, while the area integrand carries $y = 1 / x$ , whose improper integral diverges.

The Surface Area Element

Revolving About the x-Axis

Revolving About the y-Axis

Worked Examples: Revolving About the x-Axis

Worked Examples: Revolving About the y-Axis

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