Arc Length and Surface Area

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.

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.

Template:ExampleRobox 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

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

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:

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

Resources

Arc Length

• Arc Length - Part 1 of 2 by James Sousa, Math is Power 4U
• Arc Length - Part 2 of 2 by James Sousa, Math is Power 4U
• Ex: Find the Arc Length of a Linear Function by James Sousa, Math is Power 4U
• Ex: Find the Arc Length of a Radical Function by James Sousa, Math is Power 4U
• Ex: Find the Arc Length of a Quadratic Function by James Sousa, Math is Power 4U

• Deriving the Arc Length Formula in Calculus by patrickJMT
• Arc Length by patrickJMT

• Arc Length y=f(x) by Krista King
• Arc length x=g(y) by Krista King

• Arc Length Intro by Khan Academy
• Arc Length Example by Khan Academy
• Arc Length Example by Khan Academy

• Arc Length by The Organic Chemistry Tutor

Surface Area

• Surface Area of Revolution - Part 1 of 2 by James Sousa, Math is Power 4U
• Surface Area of Revolution - Part 2 of 2 by James Sousa, Math is Power 4U
• Ex: Surface Area of Revolution - Linear Function by James Sousa, Math is Power 4U
• Ex: Surface Area of Revolution - Sine Function by James Sousa, Math is Power 4U
• Ex: Surface Area of Revolution - Cubic Function About x-axis by James Sousa, Math is Power 4U
• Ex: Surface Area of Revolution - Square Root Function About x-axis by James Sousa, Math is Power 4U
• Ex: Surface Area of Revolution - Quadratic Function About y-axis by James Sousa, Math is Power 4U
• Ex: Surface Area of Revolution - Cube Root Function About y-axis by James Sousa, Math is Power 4U

• Finding Surface Area - Part 1 by patrickJMT
• Finding Surface Area - Part 2 by patrickJMT

• Surface Area of Revolution Example 1 by Krista King
• Surface Area of Revolution Example 2 by Krista King
• Surface Area of Revolution Example 3 by Krista King

• Surface Area of Revolution By Integration by The Organic Chemistry Tutor