Applied Integration - Arc Length [ Rectangular Coordinates ]

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For arc length of parametric curves, see the parametrics calculus page.

For the arc length of polar curves, see the polar calculus page.

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This page covers the topic of arc length of an explicitly defined smooth curve in the xy-plane in cartesian coordinates.

Setting up these integrals is very straight-forward and do not require any new techniques. However, the difficulty comes in evaluating them. We will show you a few tricks to evaluating them but first let's talk about how to set them up.

There are two sets of equations, both of which accomplish the same purpose. The reason you need two of them is because sometimes the equations describing the curve are in a form that require one or the other. Additionally, you may be able to set up both integrals but often it is possible to evaluate only one of them.

Here is a video clip deriving the integral in terms of x. This will help you understand where these equations come from.

To determine the arc length of a smooth curve defined by \(y = f(x) \) between the points \( x=a \) and \( x=b \), we can use the integral

When the curve is given by an equation in the form \(x=g(y)\), we need to determine the endpoints, \(y=c\) and \(y=d\) on the y-axis and use the integral

\(\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2}dx}}\)

\(\displaystyle{s = \int_{c}^{d}{\sqrt{1 + [g'(y)]^2}dy}}\)

where \(\displaystyle{ f'(x) = \frac{dy}{dx} }\)

where \(\displaystyle{ g'(y) = \frac{dx}{dy} }\)

Okay, let's watch a quick video clip going through these equations again.

Notes:
1. Notice that we use a small s to represent the arc length. Later, we will use a capital S to represent surface area. Many, if not most, mathematicians follow this standard.
2. Sometimes these equations are represented a little differently. These can be written as shown below. When you study surface area, you will see how these are used.

\(ds = \sqrt{1 + [f'(x)]^2}dx \)

\( ds = \sqrt{1 + [g'(y)]^2}dy \)

Tricks To Evaluate Integrals

Here are some unusual situation you may encounter when evaluating integrals that have not been covered elsewhere. We will show some specific examples but we will not evaluate the integrals completely. We will just get them into a form that can be evaluated with other techniques that you are already know, like substitution or parts.

Perfect Square

Okay, so I know that you already know about perfect squares but they show up a lot in these problems, so I thought I would remind you. Notice in the integral, we have \(\sqrt{1+[f'(x)]^2}\).   After expanding the squared derivative and adding one, the easiest problems end up with a perfect square under the square root. So, even with complicated expressions, it is usually a good use of your time to check for a perfect square first, even if it isn't obvious. Here is an example. After expanding the squared term and adding one, we have \(\displaystyle{\sqrt{ x^2 + \frac{1}{2} + \frac{1}{16x^2}}}\)

We can try to set up a perfect square using the square root of the first and last term and see if we get the middle term under the square root when we expand. So we investigate \(\displaystyle{ x + \frac{1}{4x} }\). When we square this, sure enough we get \(1/2\) as the middle term. So we can simplify the equation as follows
\(\displaystyle{\sqrt{ x^2 + \frac{1}{2} + \frac{1}{16x^2} } = }\) \(\displaystyle{ \sqrt{ \left[ x + \frac{1}{4x} \right]^2 } = }\) \(\displaystyle{ x + \frac{1}{4x} }\)
This last form is easily integrable with the techniques you know.

Getting A Common Denominator

Okay, so you have determined that you do not have a perfect square under the square root. Your next step might be to combine all terms into one fraction and then take the square root of the numerator and denominator separately. This will often allow you to move some terms outside of the square root and make the square root easier and may lead to integration by substitution or trig substitution. Here is an example.

Let's say we have an integral with the integrand \(\displaystyle{ \sqrt{ 1+\frac{y^2-2y+1}{4y} } }\). In order to simplify this so that we can integrate it, we need to combine the one with the fraction. Doing this we get \(\displaystyle{ \sqrt{ \frac{y^2+2y+1}{4y} } }\). Now, the numerator is a perfect square, so we factor it and simplify to get \(\displaystyle{ \frac{ \sqrt{ (y+1)^2 } }{\sqrt{4y} } =\frac{y+1}{\sqrt{4y}}}\). So, now we have an integrand that can be integrated with basic techniques.

Arc Length Resources

Practice Problems

Instructions - - Unless otherwise instructed, find the arc length. Give all answers in exact form.
Note: Although some of these problems are very similar, they are worked by different instructors. So watching each solution will give you a broader point of view and help you better understand how to work these problems.
[ Click on the practice problem to reveal/hide the solution. ]

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