series of arc sculptures representing arc length formula calculus

Calculus

The Arc-Length Formula: A Step-By-Step Guide

04.25.2022 • 8 min read

Rachel McLean

Subject Matter Expert

What is the arc length formula? First, we’ll learn how to derive the arc length formula. Then, we’ll discuss how to find the arc length and practice with some examples. Finally, we’ll apply our knowledge of the arc length formula to help us calculate the surface area of a surface of revolution.

Share

In This Article

  1. The Arc Length Formula

  2. How To Find the Arc Length of a Function

  3. What Is the Surface Area of the Surface of Revolution?

  4. How To Calculate the Surface Area of a Surface of Revolution

The Arc Length Formula

The arc length of a function is the length of the function’s curve between two points.

Graph showing the arc length of a function f on [a, b]

In order to calculate the arc length of a function ff on [a,b][a, b], we require two things: the function must be differentiable on [a,b][a, b] and its derivative must be continuous on [a,b][a, b]. Functions with these characteristics are called smooth.

Suppose that ff is smooth on [a,b][a, b]. To approximate the arc length of the curve, we can break the curve into nn small sections. Then, we can connect the endpoints of each section with a straight line to form an approximation of the curve.

Using the distance formula, we can determine the length of each straight line segment. Adding up these lengths gives us an approximate answer for the arc length. Below is an example of this process for n=5n = 5.

Graph showing straight line segments of an arc. Each straight line segment shows the hypotenuse of a right triangle.

Each straight line segment is the hypotenuse of a right triangle. Then, by the Pythagorean theorem, the length of each straight line segment is given by the formula below. This is called the distance formula.

si=(xixi1)2+(yiyi1)2s_i = \sqrt{(x_i - x_{i-1})^2 + (y_i - y_{i-1})^2}

Because the curve on [a,b][a, b] is broken into nn pieces using a regular partition over [a,b][a, b], we can let the change in horizontal distance be given by Δx\Delta x. However, the change in vertical distances varies, so this is given by Δyi=yiyi1\Delta y_i = y_i - y_{i-1}. Then, we have:

si=Δx2+Δyi2s_i = \sqrt{\Delta x^2 + \Delta y_i^2}

Taking the sum of the lengths of these tiny straight line segments gives us an approximate measurement of the arc length.

Arc Lengthi=1nΔx2+Δyi2\text{Arc Length} \approx \sum_{i = 1}^n \sqrt{\Delta x^2 + \Delta y_i^2}

Now, notice that Δyi=yiyi1=f(xi)f(xi1)\Delta y_i = y_i - y_{i-1} = f(x_i) - f(x_{i-1}). This allows us to use the Mean Value Theorem, which states that if ff is continuous and differentiable on [a,b][a, b], then there exists some point cc in [a,b][a, b] such that

f(c)=f(b)f(a)baf’(c) = \frac{f(b) - f(a)}{b-a}

So, we can say there exists some x_i^*$ in $[x_{i-1}, x_i] such that:

f(xi)=f(xi)f(xi1)xixi1f’(x_i^*) = \frac{f(x_i)-f(x_{i-1})}{x_i - x_{i-1}}

Rearranging this equation, we get:

f(xi)(xixi1)=f(xi)f(xi1)f’(x_i^*)(x_i - x_{i-1}) = f(x_i)-f(x_{i-1})

Substituting in Δyi=f(xi)f(xi1)\Delta y_i = f(x_i) - f(x_{i-1}) and Δx=xixi1\Delta x = x_i - x_{i-1}, we get:

Δyi=f(xi)Δx\Delta y_i = f’(x_i^*)\Delta x

Using this expression, the arc length can now be approximated as:

Arc Lengthi=1nΔx2+Δyi2\text{Arc Length} \approx \sum_{i = 1}^n \sqrt{\Delta x^2 + \Delta y_i^2}
i=1nΔx2+(f(xi)Δx)2\approx \sum_{i = 1}^n \sqrt{\Delta x^2 + {(f’(x_i^*) \Delta x)}^2}
i=1nΔx2+[f(xi)]2Δx2\approx \sum_{i = 1}^n \sqrt{\Delta x^2 + [f’(x_i^*)]^2 \Delta x^2}

Factoring out Δx2\Delta x^2, we have:

Arc Lengthi=1nΔx2(1+[f(xi)]2)\text{Arc Length} \approx \sum_{i = 1}^n \sqrt{\Delta x^2(1 + [f’(x_i^*)]^2)}
Arc Lengthi=1nΔx(1+[f(xi)]2)\text{Arc Length} \approx \sum_{i = 1}^n \Delta x \sqrt{(1 + [f’(x_i^*)]^2)}

As we break the curve into smaller and smaller sections, the collection of resulting straight line segments begins to match the original curve better and better.

So, as we make nn bigger, the curve of ff is broken into more and more small pieces, and our approximation of the arc length becomes more and more precise.

Then, by taking the limit of our approximation as nn approaches infinity, we can find the precise arc length of ff on [a,b][a, b].

Arc Length=limni=1nΔx1+[f(xi)]2\text{Arc Length} = \lim_{n\to\infty} \sum_{i = 1}^n \Delta x \sqrt{1 + [f’(x_i^*)]^2}

Note that a definite integral is defined by:

abf(x)dx=limni=1nf(xi)Δx\int_a^b f(x) dx = \lim_{n\to\infty} \sum_{i = 1}^n f’(x_i^*) \Delta x

Using this definition, we can say that the arc length is equal to the definite integral below. We have finally derived the arc length equation.

Arc Length=ab1+[f(x)]2dx\text{Arc Length} = \int_a^b \sqrt{1 + [f’(x)]^2} dx

Similarly, if g(y)g(y) is a smooth function on [c,d][c, d], we can say that

Arc Length=cd1+[g(y)]2dy\text{Arc Length} = \int_c^d \sqrt{1 + [g’(y)]^2} dy

How To Find the Arc Length of a Function

Now that we understand how to derive the arc length integral formula, we can follow the four simple steps below to calculate the arc length of a smooth function on [a,b][a, b].

  1. Differentiate f(x)f(x) to find f(x)f’(x).

  2. Square f(x)f’(x).

  3. Plug [f(x)]2[f’(x)]^2 into the arc length formula and plug aa and bb into the upper and lower bounds of the integral.

  4. Integrate.

How To Find an Arc Length Example

Let’s do one example together to solidify your understanding. Let f(x)=x32f(x) = x^{\frac{3}{2}}. Find the arc length of ff from x=4x = 4 to x=8x = 8.

  1. Differentiating f(x)f(x) using the power rule, we have f(x)=32x12f’(x) = \frac{3}{2}x^\frac{1}{2}.

  2. Squaring f(x)f’(x), we find that [f(x)]2=(32x12)2=94x[f’(x)]^2 = {(\frac{3}{2}x^\frac{1}{2})}^2 = \frac{9}{4}x.

  3. Using the arc length formula, our integral is 131+94xdx\int_1^3 \sqrt{1 + \frac{9}{4}x} dx.

  4. We will integrate using u-substitution. (If needed, you can review our guide about what is u-substitution.

Let u=1+94xu = 1 + \frac{9}{4}x. Then du=94dxdu = \frac{9}{4} \, dx. Then, to keep our equation balanced, we must multiply the integrand by 49\frac{4}{9}. We must also calculate our new bounds in terms of uu. Plugging a=4a = 4 into uu, we get u(4)=10u(4) = 10. Plugging b=8b = 8 into uu, we get u(8)=19u(8) = 19. Now, we can integrate. You will need to use your calculator to compute this and get the approximate final answer.

Arc Length=101949udu\text{Arc Length} = \int_{10}^{19} \frac{4}{9} \sqrt{u} \, du
=101949u12du= \int_{10}^{19} \frac{4}{9} {u}^{\frac{1}{2}} \, du
=4923u321019= \frac{4}{9} \cdot \frac{2}{3}{u}^{\frac{3}{2}} \Big|_{10}^{19}
=827u321019= \frac{8}{27}{u}^{\frac{3}{2}} \Big|_{10}^{19}
=82719328271032= \frac{8}{27}{19}^{\frac{3}{2}} - \frac{8}{27}{10}^{\frac{3}{2}}
15.1693\approx 15.1693

Thus the arc length of ff from x=4x = 4 to x=8x = 8 is approximately 15.1693.

What Is the Surface Area of the Surface of Revolution?

We can create a three dimensional object by rotating the curve of a function 360 degrees about the x-axis. This creates a surface of revolution. For example, the surface of revolution created by revolving y=x2y = x^2 about the x-axis is given below. Note that our function must be smooth and nonnegative.

Graph showing the surface of revolution created by revolving y = x^2 about the x-axis

Frustum

As we did before to derive the arc length formula, imagine breaking the curve of ff into nn small sections and connecting the endpoints of each section with a straight line segment. Revolving these straight line segments about the x-axis creates a three-dimensional shape that looks like a piece of cone called a frustum. A frustum looks like an ice cream cone with the pointy part removed.

Below is an example of a frustum generated by rotating a straight line segment around the x-axis.

Graph showing an example of a frustum generated by rotating a straight line segment around the x-axis

Formula for the Surface Area of a Frustum

So, how do we calculate the surface area of the surface of revolution? Well, we can start by investigating the formula for the surface area of a frustum.

The formula for the lateral surface area of a frustum is given by SA=2π(r1+r2)2lSA = 2\pi \frac{(r_1 + r_2)}{2}l, where r1r_1 and r2r_2 are the radii of the bases and ll is the slant height of the frustum.

The radii r1r_1 and r2r_2 are equal to the values yi=f(xi)y_i = f(x_i) and yi1=f(xi1)y_{i-1} = f(x_{i-1}), respectively. The slant height ll is simply the length of the line segment used to generate the frustum. We already calculated the formula for the length of the straight line segment in our previous work for deriving the arc length formula.

So, we can change the formula for the surface area of a frustum to look like this:

SA=2π(r1+r2)2lSA = 2\pi \frac{(r_1 + r_2)}{2}l
SA=2π(f(xi)+f(xi1)2)Δx2+Δyi2SA = 2\pi (\frac{f(x_i) + f(x_{i-1})}{2}) \sqrt{\Delta x^2 + \Delta y_i^2}
SA=2π(f(xi)+f(xi1)2)(1+[f(xi)]2)ΔxSA = 2\pi (\frac{f(x_i) + f(x_{i-1})}{2}) \sqrt{(1 + [f’(x_i^*)]^2)} \Delta x

The Intermediate Value Theorem tells us that there exists some value f(xi)f(x_i^*) between f(xi1)f(x_{i-1}) and f(xi)f(x_i) such that f(xi)=f(xi)+f(xi1)2f(x_i^*) = \frac{f(x_i) + f(x_{i-1})}{2}, so our equation becomes:

SA=2πf(xi)(1+[f(xi)]2)ΔxSA = 2\pi f(x_i^*) \sqrt{(1 + [f’(x_i^*)]^2)} \Delta x

Taking the sum of the surface area of each frustum that is generated by the nn straight line segments that approximate the curve of ff gives us:

Surface Area of the Surface of Revolutioni=1n2πf(xi)(1+[f(xi)]2)Δx\text{Surface Area of the Surface of Revolution} \approx \sum_{i = 1}^n 2\pi f(x_i^*) \sqrt{(1 + [f’(x_i^*)]^2)} \Delta x

Similar to what we determined with the arc length formula, when we break the curve into smaller and smaller sections, the collection of resulting frustums begins to match the surface of revolution better and better.

As we make nn bigger, the curve of ff is broken into more and more small pieces, and our approximation of the surface area of the surface of revolution becomes more and more precise.

Then, by taking the limit of our surface area approximation as nn approaches infinity, we can find the precise surface area of the surface of revolution of ff on [a,b][a, b].

SA of the Surface of Revolution=limni=1n2πf(xi)(1+[f(xi)]2)Δx\text{SA of the Surface of Revolution} = \lim_{n\to\infty} \sum_{i = 1}^n 2\pi f(x_i^*) \sqrt{(1 + [f’(x_i^*)]^2)} \Delta x

We can use the definition of a definite integral that was given before to finally determine the formula for the surface area of a surface of revolution given by revolving ff around the x-axis on [a,b][a, b].

SA of the Surface of Revolution=ab2πf(x)1+[f(x)]2dx\text{SA of the Surface of Revolution} = \int_a^b 2\pi f(x) \sqrt{1 + [f’(x)]^2} dx

Similarly, the surface area of the surface of revolution given by revolving a nonnegative smooth function gg around the y-axis on [c,d][c, d] is:

SA of the Surface of Revolution=cd2πg(y)1+[g(y)]2dy\text{SA of the Surface of Revolution} = \int_c^d 2\pi g(y) \sqrt{1 + [g’(y)]^2} dy

How To Calculate the Surface Area of a Surface of Revolution

Now that we understand how to derive the formula, we can follow the four simple steps below to calculate the surface area of the surface of revolution of a smooth function on [a,b][a, b].

  1. Differentiate f(x)f(x) to find f(x)f’(x).

  2. Square f(x)f’(x).

  3. Plug [f(x)]2[f’(x)]^2 into the surface area formula and plug aa and bb into the upper and lower bounds of the integral.

  4. Integrate.

Example of How To Find the Surface Area of the Surface of Revolution

We’ll do one simple example together. Let f(x)=xf(x) = x. Find the surface area of the surface of revolution on [0,1][0, 1] formed by revolving the graph of f(x)f(x) around the x-axis.

  1. Differentiating f(x)f(x) using the power rule, we find the f(x)=1f’(x) = 1.

  2. Squaring f(x)f’(x) gives us 1.

  3. Using the surface area formula, our integral is:

SA of the Surface of Revolution=012πx1+1dx\text{SA of the Surface of Revolution} = \int_0^1 2\pi x \sqrt{1 + 1} dx
=012πx2dx= \int_0^1 2\pi x \sqrt{2} dx
=0122πxdx= \int_0^1 2 \sqrt{2} \pi x dx
=22π01xdx=2 \sqrt{2} \pi \int_0^1 x dx

4. Integrating gives us:

22π01xdx=22πx22012 \sqrt{2} \pi \int_0^1 x dx = 2 \sqrt{2} \pi \cdot \frac{x^2}{2} \Big|_0^1
=22π(120) = 2 \sqrt{2} \pi (\frac{1}{2}-0)
=π2 = \pi \sqrt{2}

Thus, the surface area of the surface of revolution of f(x)=xf(x) = x on [0,1][0, 1] is π24.4429\pi \sqrt{2} \approx 4.4429.

Explore Outlier's Award-Winning For-Credit Courses

Outlier (from the co-founder of MasterClass) has brought together some of the world's best instructors, game designers, and filmmakers to create the future of online college.

Check out these related courses:

Calculus I

Calculus I

Explore course

Intro to Statistics

Intro to Statistics

Explore course

Intro to Microeconomics

Intro to Microeconomics

Explore course