Arc Length and Surface Area

Arc Length

Suppose that we are given a function $f$ that is continuous on an interval $[a,b]$ and we want to calculate the length of the curve drawn out by the graph of $f(x)$ from $x=a$ to $x=b$ . If the graph were a straight line this would be easy — the formula for the length of the line is given by Pythagoras' theorem. And if the graph were a piecewise linear function we can calculate the length by adding up the length of each piece.

The problem is that most graphs are not linear. Nevertheless we can estimate the length of the curve by approximating it with straight lines. Suppose the curve $C$ is given by the formula $y=f(x)$ for $a\leq x\leq b$ . We divide the interval $[a,b]$ into $n$ subintervals with equal width $\Delta x$ and endpoints $x_{0},x_{1},\ldots ,x_{n}$ . Now let $y_{i}=f(x_{i})$ so $P_{i}=(x_{i},y_{i})$ is the point on the curve above $x_{i}$ . The length of the straight line between $P_{i}$ and $P_{i+1}$ is

{\bigl |}P_{i}P_{i+1}{\bigr |}={\sqrt {(y_{i+1}-y_{i})^{2}+(x_{i+1}-x_{i})^{2}}}

So an estimate of the length of the curve $C$ is the sum

\sum _{i=0}^{n-1}{\bigl |}P_{i}P_{i+1}{\bigr |}

As we divide the interval $[a,b]$ into more pieces this gives a better estimate for the length of $C$ . In fact we make that a definition.

Length of a Curve'

The length of the curve

y=f(x)

for

a\leq x\leq b

is defined to be

L=\lim _{n\to \infty }\sum _{i=0}^{n-1}{\bigl |}P_{i+1}P_{i}{\bigr |}

The Arclength Formula

Suppose that $f'$ is continuous on $[a,b]$ . Then the length of the curve given by $y=f(x)$ between $a$ and $b$ is given by

L=\int \limits _{a}^{b}{\sqrt {1+f'(x)^{2}}}dx

And in Leibniz notation

L=\int \limits _{a}^{b}{\sqrt {1+\left({\tfrac {dy}{dx}}\right)^{2}}}dx

Proof: Consider $y_{i+1}-y_{i}=f(x_{i+1})-f(x_{i})$ . By the Mean Value Theorem there is a point $z_{i}$ in $(x_{i+1},x_{i})$ such that

y_{i+1}-y_{i}=f(x_{i+1})-f(x_{i})=f'(z_{i})(x_{i+1}-x_{i})

So

${\bigl \|}P_{i}P_{i+1}{\bigr \|}$	$={\sqrt {(x_{i+1}-x_{i})^{2}+(y_{i+1}-y_{i})^{2}}}$
	$={\sqrt {(x_{i+1}-x_{i})^{2}+f'(z_{i})^{2}(x_{i+1}-x_{i})^{2}}}$
	$={\sqrt {{\bigl (}1+f'(z_{i})^{2}{\bigr )}(x_{i+1}-x_{i})^{2}}}$
	$={\sqrt {1+f'(z_{i})^{2}}}\Delta x$

Putting this into the definition of the length of $C$ gives

L=\lim _{n\to \infty }\sum _{i=0}^{n-1}{\sqrt {1+f'(z_{i})^{2}}}\Delta x

Now this is the definition of the integral of the function $g(x)={\sqrt {1+f'(x)^{2}}}$ between $a$ and $b$ (notice that $g$ is continuous because we are assuming that $f'$ is continuous). Hence

L=\int \limits _{a}^{b}{\sqrt {1+f'(x)^{2}}}dx

as claimed.

Example

Length of the curve

y=2x

from

x=0

to

x=1

}}

As a sanity check of our formula, let's calculate the length of the "curve" $y=2x$ from $x=0$ to $x=1$ . First let's find the answer using the Pythagorean Theorem.

P_{0}=(0,0)

and

P_{1}=(1,2)

so the length of the curve, $s$ , is

s={\sqrt {2^{2}+1^{2}}}={\sqrt {5}}

Now let's use the formula

s=\int \limits _{0}^{1}{\sqrt {1+\left({\tfrac {d(2x)}{dx}}\right)^{2}}}\,dx=\int \limits _{0}^{1}{\sqrt {1+2^{2}}}\,dx={\sqrt {5}}x{\bigg |}_{0}^{1}={\sqrt {5}}

Exercises

1. Find the length of the curve $y=x{\sqrt {x}}$ from $x=0$ to $x=1$ .

2. Find the length of the curve $y={\frac {e^{x}+e^{-x}}{2}}$ from $x=0$ to $x=1$ .

Arclength of a parametric curve

For a parametric curve, that is, a curve defined by $x=f(t)$ and $y=g(t)$ , the formula is slightly different:

L=\int \limits _{a}^{b}{\sqrt {f'(t)^{2}+g'(t)^{2}}}\,dt

Proof: The proof is analogous to the previous one: Consider $y_{i+1}-y_{i}=g(t_{i+1})-g(t_{i})$ and $x_{i+1}-x_{i}=f(t_{i+1})-f(t_{i})$ .

By the Mean Value Theorem there are points $c_{i}$ and $d_{i}$ in $(t_{i+1},t_{i})$ such that

y_{i+1}-y_{i}=g(t_{i+1})-g(t_{i})=g'(c_{i})(t_{i+1}-t_{i})

and

x_{i+1}-x_{i}=f(t_{i+1})-f(t_{i})=f'(d_{i})(t_{i+1}-t_{i})

So

${\bigl \|}P_{i}P_{i+1}{\bigr \|}$	$={\sqrt {(x_{i+1}-x_{i})^{2}+(y_{i+1}-y_{i})^{2}}}$
	$={\sqrt {f'(d_{i})^{2}(t_{i+1}-t_{i})^{2}+g'(c_{i})^{2}(t_{i+1}-t_{i})^{2}}}$
	$={\sqrt {{\bigl (}f'(d_{i})^{2}+g'(c_{i})^{2}{\bigr )}(t_{i+1}-t_{i})^{2}}}$
	$={\sqrt {f'(d_{i})^{2}+g'(c_{i})^{2}}}\Delta t$

Putting this into the definition of the length of the curve gives

L=\lim _{n\to \infty }\sum _{i=0}^{n-1}{\sqrt {f'(d_{i})^{2}+g'(c_{i})^{2}}}\Delta t

This is equivalent to:

L=\int \limits _{a}^{b}{\sqrt {f'(t)^{2}+g'(t)^{2}}}\,dt

Exercises

3. Find the circumference of the circle given by the parametric equations $x(t)=R\cos(t)$ , $y(t)=R\sin(t)$ , with $t$ running from $0$ to $2\pi$ .

4. Find the length of one arch of the cycloid given by the parametric equations $x(t)=R{\bigl (}t-\sin(t){\bigr )}$ , $y(t)=R{\bigl (}1-\cos(t){\bigr )}$ , with $t$ running from $0$ to $2\pi$ .

Exercise Solutions

${\frac {13{\sqrt {13}}-8}{27}}$
${\frac {e-{\frac {1}{e}}}{2}}$
$2\pi R$
$8R$

Surface Area

Suppose we are given a function $f$ and we want to calculate the surface area of the function $f$ rotated around a given line. The calculation of surface area of revolution is related to the arc length calculation.

If the function $f$ is a straight line, other methods such as surface area formulae for cylinders and conical frusta can be used. However, if $f$ is not linear, an integration technique must be used.

Recall the formula for the lateral surface area of a conical frustum:

A=2\pi rl

where $r$ is the average radius and $l$ is the slant height of the frustum.

For $y=f(x)$ and $a\leq x\leq b$ , we divide $[a,b]$ into subintervals with equal width $\delta x$ and endpoints $x_{0},x_{1},\ldots ,x_{n}$ . We map each point $y_{i}=f(x_{i})$ to a conical frustum of width Δx and lateral surface area $A_{i}$ .

We can estimate the surface area of revolution with the sum

A=\sum _{i=0}^{n}A_{i}

As we divide $[a,b]$ into smaller and smaller pieces, the estimate gives a better value for the surface area.

Definition (Surface of Revolution)

The surface area of revolution of the curve $y=f(x)$ about a line for $a\leq x\leq b$ is defined to be

$A=\lim _{n\to \infty }\sum _{i=0}^{n}A_{i}$

The Surface Area Formula

Suppose $f$ is a continuous function on the interval $[a,b]$ and $r(x)$ represents the distance from $f(x)$ to the axis of rotation. Then the lateral surface area of revolution about a line is given by

A=2\pi \int _{a}^{b}r(x){\sqrt {1+f'(x)^{2}}}\,dx

And in Leibniz notation

A=2\pi \int _{a}^{b}r(x){\sqrt {1+\left({\tfrac {dy}{dx}}\right)^{2}}}\,dx

Proof:

$A$	$=\lim _{n\to \infty }\sum _{i=1}^{n}A_{i}$
	$=\lim _{n\to \infty }\sum _{i=1}^{n}2\pi r_{i}l_{i}$
	$=2\pi \cdot \lim _{n\to \infty }\sum _{i=1}^{n}r_{i}l_{i}$

As $n\to \infty$ and $\Delta x\to 0$ , we know two things:

the average radius of each conical frustum $r_{i}$ approaches a single value
the slant height of each conical frustum $l_{i}$ equals an infitesmal segment of arc length

From the arc length formula discussed in the previous section, we know that

l_{i}={\sqrt {1+f'(x_{i})^{2}}}

Therefore

$A$	$=2\pi \cdot \lim _{n\to \infty }\sum _{i=1}^{n}r_{i}l_{i}$
	$=2\pi \cdot \lim _{n\to \infty }\sum _{i=1}^{n}r_{i}{\sqrt {1+f'(x_{i})^{2}}}\Delta x$

Because of the definition of an integral $\int _{a}^{b}f(x)dx=\lim _{n\to \infty }\sum _{i=1}^{n}f(c_{i})\Delta x_{i}$ , we can simplify the sigma operation to an integral.