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Arc Length of a Curve and Surface Area

Determine the length of a curve,

Arc Length of the Curve

In previous applications of integration, we required the function \(f(x)\) to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for \(f(x).\) Here, we require \(f(x)\) to be differentiable, and furthermore we require its derivative, \({f}^{'}(x),\) to be continuous. Functions like this, which have continuous derivatives, are called smooth. (This property comes up again in later chapters.)

Let \(f(x)\) be a smooth function defined over \([a,b].\) We want to calculate the length of the curve from the point \((a,f(a))\) to the point \((b,f(b)).\) We start by using line segments to approximate the length of the curve. For \(i=0,\ 1,2\text{,\ldots },n,\) let \(P=\{{x}_{i}\}\) be a regular partition of \([a,b].\) Then, for \(i=1,2\text{,\ldots },n,\) construct a line segment from the point \(({x}_{i-1},f({x}_{i-1}))\) to the point \(({x}_{i},f({x}_{i})).\) Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. depicts this construct for \(n=5.\)

To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by \(\text{\Delta }x.\) The change in vertical distance varies from interval to interval, though, so we use \(\text{\Delta }{y}_{i}=f({x}_{i})-f({x}_{i-1})\) to represent the change in vertical distance over the interval \([{x}_{i-1},{x}_{i}],\) as shown in . Note that some (or all) \(\text{\Delta }{y}_{i}\) may be negative.

By the Pythagorean theorem, the length of the line segment is \(\sqrt{{(\text{\Delta }x)}^{2}+{(\text{\Delta }{y}_{i})}^{2}}.\) We can also write this as \(\text{\Delta }x\sqrt{1+{((\text{\Delta }{y}_{i})\text{/}(\text{\Delta }x))}^{2}}.\) Now, by the Mean Value Theorem, there is a point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}]\) such that \({f}^{'}({x}_{i}^{*})=(\text{\Delta }{y}_{i})\text{/}(\text{\Delta }x).\) Then the length of the line segment is given by \(\text{\Delta }x\sqrt{1+{[{f}^{'}({x}_{i}^{*})]}^{2}}.\) Adding up the lengths of all the line segments, we get

\[\text{Arc Length}\ \approx \sum _{i=1}^{n}\sqrt{1+{[{f}^{'}({x}_{i}^{*})]}^{2}}\ \text{\Delta }x.\]

This is a Riemann sum. Taking the limit as \(n\to \infty ,\) we have

\[\text{Arc Length}=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\sqrt{1+{[{f}^{'}({x}_{i}^{*})]}^{2}}\ \text{\Delta }x={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx.\]

We summarize these findings in the following theorem.

Condensed — the full section is in OpenStax Calculus Volume 2.

Arc Length of the Curve

We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of \(y,\) we can repeat the same process, except we partition the \(y\text{-axis}\) instead of the \(x\text{-axis}.\) shows a representative line segment.

Then the length of the line segment is \(\sqrt{{(\text{\Delta }y)}^{2}+{(\text{\Delta }{x}_{i})}^{2}},\) which can also be written as \(\text{\Delta }y\sqrt{1+{((\text{\Delta }{x}_{i})\text{/}(\text{\Delta }y))}^{2}}.\) If we now follow the same development we did earlier, we get a formula for arc length of a function \(x=g(y).\)

Example

Try it.

Let \(g(y)=3{y}^{3}.\) Calculate the arc length of the graph of \(g(y)\) over the interval \([1,2].\)

Solution

We have \({g}^{'}(y)=9{y}^{2},\) so \({[{g}^{'}(y)]}^{2}=81{y}^{4}.\) Then the arc length is

\[\text{Arc Length}={\int }_{c}^{d}\sqrt{1+{[{g}^{'}(y)]}^{2}}\ dy={\int }_{1}^{2}\sqrt{1+81{y}^{4}}\ dy.\]

Using a computer to approximate the value of this integral, we obtain

\[{\int }_{1}^{2}\sqrt{1+81{y}^{4}}\ dy\approx 21.0277.\]

Area of a Surface of Revolution

The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. For curved surfaces, the situation is a little more complex. Let \(f(x)\) be a nonnegative smooth function over the interval \([a,b].\) We wish to find the surface area of the surface of revolution created by revolving the graph of \(y=f(x)\) around the \(x\text{-axis}\) as shown in the following figure.

As we have done many times before, we are going to partition the interval \([a,b]\) and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For \(i=0,1,2\text{,\ldots },n,\) let \(P=\{{x}_{i}\}\) be a regular partition of \([a,b].\) Then, for \(i=1,2\text{,\ldots },n,\) construct a line segment from the point \(({x}_{i-1},f({x}_{i-1}))\) to the point \(({x}_{i},f({x}_{i})).\) Now, revolve these line segments around the \(x\text{-axis}\) to generate an approximation of the surface of revolution as shown in the following figure.

Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones (think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone.

To find the surface area of the band, we need to find the lateral surface area, \(S,\) of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). Let \({r}_{1}\) and \({r}_{2}\) be the radii of the wide end and the narrow end of the frustum, respectively, and let \(l\) be the slant height of the frustum as shown in the following figure.

We know the lateral surface area of a cone is given by

\[\text{Lateral Surface Area}=\pi rs,\]

where \(r\) is the radius of the base of the cone and \(s\) is the slant height (see the following figure).

Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surface area of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (see the following figure).

\[\frac{{r}_{2}}{{r}_{1}}=\frac{s-l}{s}.\]\[\begin{array}{lll}\frac{{r}_{2}}{{r}_{1}} & = & \frac{s-l}{s} \\ {r}_{2}s & = & {r}_{1}(s-l) \\ {r}_{2}s & = & {r}_{1}s-{r}_{1}l \\ {r}_{1}l & = & {r}_{1}s-{r}_{2}s \\ {r}_{1}l & = & ({r}_{1}-{r}_{2})s\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

The arc length of a curve can be calculated using a definite integral.
The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then gives us the definite integral formula. The same process can be applied to functions of \(y.\)
The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution.
The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. It may be necessary to use a computer or calculator to approximate the values of the integrals.

Key Equations

Arc Length of a Function of x	\(\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx\)
Arc Length of a Function of y	\(\text{Arc Length}={\int }_{c}^{d}\sqrt{1+{[{g}^{'}(y)]}^{2}}\ dy\)
Surface Area of a Function of x	\(\text{Surface Area}={\int }_{a}^{b}(2\pi f(x)\sqrt{1+{({f}^{'}(x))}^{2}})dx\)

Arc Length of a Curve and Surface Area

For the following exercises, find the length of the functions over the given interval.

For the following exercises, find the lengths of the functions of \(x\) over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

For the following exercises, find the lengths of the functions of \(y\) over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

For the following exercises, find the surface area of the volume generated when the following curves revolve around the \(x\text{-axis}.\) If you cannot evaluate the integral exactly, use your calculator to approximate it.

For the following exercises, find the surface area of the volume generated when the following curves revolve around the \(y\text{-axis}\text{.}\) If you cannot evaluate the integral exactly, use your calculator to approximate it.

For the following exercises, find the exact arc length for the following problems over the given interval.

Arc Length of the Curve

\[\text{Arc Length}\ \approx \sum _{i=1}^{n}\sqrt{1+{[{f}^{'}({x}_{i}^{*})]}^{2}}\ \text{\Delta }x.\]

This is a Riemann sum. Taking the limit as \(n\to \infty ,\) we have

\[\text{Arc Length}=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\sqrt{1+{[{f}^{'}({x}_{i}^{*})]}^{2}}\ \text{\Delta }x={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx.\]

We summarize these findings in the following theorem.

Condensed — the full section is in OpenStax Calculus Volume 1.

Arc Length of the Curve

Example

Try it.

Let \(g(y)=3{y}^{3}.\) Calculate the arc length of the graph of \(g(y)\) over the interval \([1,2].\)

Solution

We have \({g}^{'}(y)=9{y}^{2},\) so \({[{g}^{'}(y)]}^{2}=81{y}^{4}.\) Then the arc length is

\[\text{Arc Length}={\int }_{c}^{d}\sqrt{1+{[{g}^{'}(y)]}^{2}}\ dy={\int }_{1}^{2}\sqrt{1+81{y}^{4}}\ dy.\]

Using a computer to approximate the value of this integral, we obtain

\[{\int }_{1}^{2}\sqrt{1+81{y}^{4}}\ dy\approx 21.0277.\]

Area of a Surface of Revolution

We know the lateral surface area of a cone is given by

\[\text{Lateral Surface Area}=\pi rs,\]

where \(r\) is the radius of the base of the cone and \(s\) is the slant height (see the following figure).

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

The arc length of a curve can be calculated using a definite integral.
The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then gives us the definite integral formula. The same process can be applied to functions of \(y.\)
The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution.
The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. It may be necessary to use a computer or calculator to approximate the values of the integrals.

Key Equations

Arc Length of a Function of x	\(\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx\)
Arc Length of a Function of y	\(\text{Arc Length}={\int }_{c}^{d}\sqrt{1+{[{g}^{'}(y)]}^{2}}\ dy\)
Surface Area of a Function of x	\(\text{Surface Area}={\int }_{a}^{b}(2\pi f(x)\sqrt{1+{({f}^{'}(x))}^{2}})dx\)

Arc Length of a Curve and Surface Area

For the following exercises, find the length of the functions over the given interval.

For the following exercises, find the lengths of the functions of \(x\) over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

For the following exercises, find the lengths of the functions of \(y\) over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

For the following exercises, find the exact arc length for the following problems over the given interval.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Let \(f(x)=2{x}^{3\text{/}2}.\) Calculate the arc length of the graph of \(f(x)\) over the interval \([0,1].\) Round the answer to three decimal places.

Bonisa impendulo

We have \({f}^{'}(x)=3{x}^{1\text{/}2},\) so \({[{f}^{'}(x)]}^{2}=9x.\) Then, the arc length is
\[\begin{array}{ll}\text{Arc Length} & ={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx \\ & ={\int }_{0}^{1}\sqrt{1+9x}\ dx.\end{array}\]
Substitute \(u=1+9x.\) Then, \(du=9\ dx.\) When \(x=0,\) then \(u=1,\) and when \(x=1,\) then \(u=10.\) Thus,
\[\begin{array}{ll}\text{Arc Length} & ={\int }_{0}^{1}\sqrt{1+9x}\ dx \\ & =\frac{1}{9}{\int }_{0}^{1}\sqrt{1+9x}9dx=\frac{1}{9}{\int }_{1}^{10}\sqrt{u}\ du \\ & ={\frac{1}{9}\cdot \frac{2}{3}{u}^{3\text{/}2}|}_{1}^{10}=\frac{2}{27}[10\sqrt{10}-1]\approx 2.268\ \text{units}.\end{array}\]
Let \(f(x)=(4\text{/}3){x}^{3\text{/}2}.\) Calculate the arc length of the graph of \(f(x)\) over the interval \([0,1].\) Round the answer to three decimal places.

Bonisa impendulo

\(\frac{1}{6}(5\sqrt{5}-1)\approx 1.697\)
Let \(f(x)={x}^{2}.\) Calculate the arc length of the graph of \(f(x)\) over the interval \([1,3].\)

Bonisa impendulo

We have \({f}^{'}(x)=2x,\) so \({[{f}^{'}(x)]}^{2}=4{x}^{2}.\) Then the arc length is given by
\[\text{Arc Length}={\int }_{a}^{b}\sqrt{1+{[{f}^{'}(x)]}^{2}}\ dx={\int }_{1}^{3}\sqrt{1+4{x}^{2}}\ dx.\]
Using a computer to approximate the value of this integral, we get
\[{\int }_{1}^{3}\sqrt{1+4{x}^{2}}\ dx\approx 8.26815.\]
Let \(f(x)=\text{sin}\ x.\) Calculate the arc length of the graph of \(f(x)\) over the interval \([0,\pi ].\) Use a computer or calculator to approximate the value of the integral.

Bonisa impendulo

\(\text{Arc Length}\approx 3.8202\)
Let \(g(y)=3{y}^{3}.\) Calculate the arc length of the graph of \(g(y)\) over the interval \([1,2].\)

Bonisa impendulo

We have \({g}^{'}(y)=9{y}^{2},\) so \({[{g}^{'}(y)]}^{2}=81{y}^{4}.\) Then the arc length is
\[\text{Arc Length}={\int }_{c}^{d}\sqrt{1+{[{g}^{'}(y)]}^{2}}\ dy={\int }_{1}^{2}\sqrt{1+81{y}^{4}}\ dy.\]
Using a computer to approximate the value of this integral, we obtain
\[{\int }_{1}^{2}\sqrt{1+81{y}^{4}}\ dy\approx 21.0277.\]
Let \(g(y)=1\text{/}y.\) Calculate the arc length of the graph of \(g(y)\) over the interval \([1,4].\) Use a computer or calculator to approximate the value of the integral.

Bonisa impendulo

\(\text{Arc Length}=3.15018\)
Let \(f(x)=\sqrt{x}\) over the interval \([1,4].\) Find the surface area of the surface generated by revolving the graph of \(f(x)\) around the \(x\text{-axis}.\) Round the answer to three decimal places.

Bonisa impendulo

The graph of \(f(x)\) and the surface of rotation are shown in the following figure.
We have \(f(x)=\sqrt{x}.\) Then, \({f}^{'}(x)=1\text{/}(2\sqrt{x})\) and \({({f}^{'}(x))}^{2}=1\text{/}(4x).\) Then,
\[\begin{array}{ll}\text{Surface Area} & ={\int }_{a}^{b}(2\pi f(x)\sqrt{1+{({f}^{'}(x))}^{2}})dx \\ & ={\int }_{1}^{4}(2\pi \sqrt{x}\sqrt{1+\frac{1}{4x}})dx \\ & ={\int }_{1}^{4}(2\pi \sqrt{x+\frac{1}{4}})dx.\end{array}\]
Let \(u=x+1\text{/}4.\) Then, \(du=dx.\) When \(x=1,\) \(u=5\text{/}4,\) and when \(x=4,\) \(u=17\text{/}4.\) This gives us
\[\begin{array}{ll}{\int }_{1}^{4}(2\pi \sqrt{x+\frac{1}{4}})dx & ={\int }_{5\text{/}4}^{17\text{/}4}2\pi \sqrt{u}\ du \\ & =2\pi {[\frac{2}{3}{u}^{3\text{/}2}]\ |}_{5\text{/}4}^{17\text{/}4}=\frac{\pi }{6}[17\sqrt{17}-5\sqrt{5}]\approx 30.846.\end{array}\]
Let \(f(x)=\sqrt{1-x}\) over the interval \([0,1\text{/}2].\) Find the surface area of the surface generated by revolving the graph of \(f(x)\) around the \(x\text{-axis}.\) Round the answer to three decimal places.

Bonisa impendulo

\(\frac{\pi }{6}(5\sqrt{5}-3\sqrt{3})\approx 3.133\)
Let \(f(x)=y=\sqrt[3]{3x}.\) Consider the portion of the curve where \(0\le y\le 2.\) Find the surface area of the surface generated by revolving the graph of \(f(x)\) around the \(y\text{-axis}.\)

Bonisa impendulo

Notice that we are revolving the curve around the \(y\text{-axis},\) and the interval is in terms of \(y,\) so we want to rewrite the function as a function of y. We get \(x=g(y)=(1\text{/}3){y}^{3}.\) The graph of \(g(y)\) and the surface of rotation are shown in the following figure.
We have \(g(y)=(1\text{/}3){y}^{3},\) so \({g}^{'}(y)={y}^{2}\) and \({({g}^{'}(y))}^{2}={y}^{4}.\) Then
\[\begin{array}{ll}\text{Surface Area} & ={\int }_{c}^{d}(2\pi g(y)\sqrt{1+{({g}^{'}(y))}^{2}})dy \\ & ={\int }_{0}^{2}(2\pi (\frac{1}{3}{y}^{3})\sqrt{1+{y}^{4}})dy \\ & =\frac{2\pi }{3}{\int }_{0}^{2}({y}^{3}\sqrt{1+{y}^{4}})dy.\end{array}\]
Let \(u={y}^{4}+1.\) Then \(du=4{y}^{3}dy.\) When \(y=0,\) \(u=1,\) and when \(y=2,\) \(u=17.\) Then
\[\begin{array}{ll}\frac{2\pi }{3}{\int }_{0}^{2}({y}^{3}\sqrt{1+{y}^{4}})dy & =\frac{2\pi }{3}{\int }_{1}^{17}\frac{1}{4}\sqrt{u}du \\ & =\frac{\pi }{6}{[\frac{2}{3}{u}^{3\text{/}2}]\ |}_{1}^{17}=\frac{\pi }{9}[{(17)}^{3\text{/}2}-1]\approx 24.118.\end{array}\]
Let \(g(y)=\sqrt{9-{y}^{2}}\) over the interval \(y\in [0,2].\) Find the surface area of the surface generated by revolving the graph of \(g(y)\) around the \(y\text{-axis}.\)

Bonisa impendulo

\(12\pi\)
\(y=5x\ \text{from}\ x=0\ \text{to}\ x=2\)

Bonisa impendulo

\(2\sqrt{26}\)
\(y=-\frac{1}{2}x+25\ \text{from}\ x=1\ \text{to}\ x=4\)
\(x=4y\ \text{from}\ y=-1\ \text{to}\ y=1\)

Bonisa impendulo

\(2\sqrt{17}\)
Pick an arbitrary linear function \(x=g(y)\) over any interval of your choice \(({y}_{1},{y}_{2}).\) Determine the length of the function and then prove the length is correct by using geometry.
Find the surface area of the volume generated when the curve \(y=\sqrt{x}\) revolves around the \(x\text{-axis}\) from \((1,1)\) to \((4,2),\) as seen here.

Bonisa impendulo

\(\frac{\pi }{6}(17\sqrt{17}-5\sqrt{5})\)
Find the surface area of the volume generated when the curve \(y={x}^{2}\) revolves around the \(y\text{-axis}\) from \((1,\ 1)\) to \((3,9).\)
\(y={x}^{3\text{/}2}\) from \((0,0)\ \text{to}\ (1,1)\)

Bonisa impendulo

\(\frac{13\sqrt{13}-8}{27}\)
\(y={x}^{2\text{/}3}\) from \((1,1)\ \text{to}\ (8,4)\)
\(y=\frac{1}{3}{({x}^{2}+2)}^{3\text{/}2}\) from \(x=0\ \text{to}\ x=1\)

Bonisa impendulo

\(\frac{4}{3}\)
\(y=\frac{1}{3}{({x}^{2}-2)}^{3\text{/}2}\) from \(x=2\) to \(x=4\)
[T] \(y={e}^{x}\) on \(x=0\) to \(x=1\)

Bonisa impendulo

\(2.0035\)
\(y=\frac{{x}^{3}}{3}+\frac{1}{4x}\) from \(x=1\ \text{to}\ x=3\)
\(y=\frac{{x}^{4}}{4}+\frac{1}{8{x}^{2}}\) from \(x=1\ \text{to}\ x=2\)

Bonisa impendulo

\(\frac{123}{32}\)
\(y=\frac{2{x}^{3\text{/}2}}{3}-\frac{{x}^{1\text{/}2}}{2}\) from \(x=1\ \text{to}\ x=4\)
\(y=\frac{1}{27}{(9{x}^{2}+6)}^{3\text{/}2}\) from \(x=0\ \text{to}\ x=2\)

Bonisa impendulo

\(10\)
[T] \(y=\text{sin}\ x\) on \(x=0\ \text{to}\ x=\pi\)
\(y=\frac{5-3x}{4}\) from \(y=0\) to \(y=4\)

Bonisa impendulo

\(\frac{20}{3}\)
\(x=\frac{1}{2}({e}^{y}+{e}^{\text{-}y})\) from \(y=-1\ \text{to}\ y=1\)
\(x=5{y}^{3\text{/}2}\) from \(y=0\) to \(y=1\)

Bonisa impendulo

\(\frac{1}{675}(229\sqrt{229}-8)\)
[T] \(x={y}^{2}\) from \(y=0\) to \(y=1\)
\(x=\sqrt{y}\) from \(y=0\ \text{to}\ y=1\)

Bonisa impendulo

\(\frac{1}{8}(4\sqrt{5}+\text{ln}(9+4\sqrt{5}))\)
\(x=\frac{2}{3}{({y}^{2}+1)}^{3\text{/}2}\) from \(y=1\) to \(y=3\)
[T] \(x=\text{tan}\ y\) from \(y=0\) to \(y=\frac{3}{4}\)

Bonisa impendulo

\(1.201\)
[T] \(x={\text{cos}}^{2}y\) from \(y=-\frac{\pi }{2}\) to \(y=\frac{\pi }{2}\)
[T] \(x={4}^{y}\) from \(y=0\ \text{to}\ y=2\)

Bonisa impendulo

\(15.2341\)
[T] \(x=\text{ln}(y)\) on \(y=\frac{1}{e}\) to \(y=e\)
\(y=\sqrt{x}\) from \(x=2\) to \(x=6\)

Bonisa impendulo

\(\frac{49\pi }{3}\)
\(y={x}^{3}\) from \(x=0\) to \(x=1\)
\(y=7x\) from \(x=-1\ \text{to}\ x=1\)

Bonisa impendulo

\(70\pi \sqrt{2}\)
[T] \(y=\frac{1}{{x}^{2}}\) from \(x=1\ \text{to}\ x=3\)

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

imaginary unit
i² = −1.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

How to: Arc Length of a Curve and Surface Area

Determine the length of a curve,
Determine the length of a curve,
Find the surface area of a solid of revolution.
The arc length of a curve can be calculated using a definite integral.
The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then gives us the definite integral formula. The same process can be applied to functions of
The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution.
The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. It may be necessary to use a computer or calculator to approximate the values of the integrals.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Zama ngokwakho

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests