This page covers the topic of arc length of an explicitly defined smooth curve in the xyplane in cartesian (rectangular) coordinates. [Arc length can also be calculated in polar coordinates and for parametric curves.]
Setting Up The Integral
Setting up these integrals is very straightforward 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.
video by MIP4U 

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

where \(\displaystyle{ f'(x) = \frac{dy}{dx} }\) 
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 yaxis and use the integral 
\(\displaystyle{s = \int_{c}^{d}{\sqrt{1 + [g'(y)]^2}dy}}\) 

where \(\displaystyle{ g'(y) = \frac{dx}{dy} }\) 
Okay, let's watch a quick video clip going through these equations again.
video by PatrickJMT 

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. We may write the arc length integral as \(s=\int{ds}\), where \(ds\) is defined 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 \(\displaystyle{ \sqrt{ 1+\frac{y^22y+1}{4y} } }\). 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 \(\displaystyle{ \frac{ \sqrt{ (y+1)^2 } }{\sqrt{4y} } =\frac{y+1}{2\sqrt{y}}}\). So, now we have an integrand that can be integrated with basic techniques.
Practice
Unless otherwise instructed, calculate 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 going through each solution will give you a broader point of view and help you better understand how to work these types problems.
\( y = 3x+2 \); \( 1 \leq x \leq 2 \)
Problem Statement 

Calculate the length of the arc \( y = 3x+2 \) for \( 1 \leq x \leq 2 \). Give your answer in exact form.
Final Answer 

\(s = 3\sqrt{10}\)
Problem Statement 

Calculate the length of the arc \( y = 3x+2 \) for \( 1 \leq x \leq 2 \). Give your answer in exact form.
Solution 

built with GeoGebra 

This is a pretty easy one where the 'arc' is a straight line, shown in the plot. We can therefore calculate the arc length directly using geometry to check our answer to the integral solution. Let's work the integral first using \(\displaystyle{s = \int_{a}^{b}{ \sqrt{1 + [f'(x)]^2} ~dx } }\)
\(f(x)=3x+2 \to f'(x)=3\) 
\(\displaystyle{s = \int_{1}^{2}{ \sqrt{1 + [3]^2} ~dx } }\) 
\(s = [ x\sqrt{10} ]_{1}^{2} = 3\sqrt{10} \) 
Double Check
Now, let's check our answer using the Pythagorean Theorem. The xlength is \(2(1) = 3\). The ylength is \(8(1) = 9\). So the length of the hypotenuse is \( s = \sqrt{ 3^2 + 9^2} = \sqrt{ 9 + 81 } = \) \( \sqrt{90} = \sqrt{9(10)} = 3\sqrt{10} \), which confirms our answer from the integral.
Triple Check
Although we were not asked to in the problem, let's evaluate the integral in the other direction, i.e. with respect to y using the integral \(\displaystyle{s = \int_{c}^{d}{ \sqrt{1 + [g'(y)]^2} ~dy } }\)
We should get the same answer.
\(f(x)=3x+2 \to \) \( g(y)=(y2)/3 \to g'(y)=1/3\) 
\(\displaystyle{s = \int_{1}^{8}{ \sqrt{1 + [1/3]^2} ~dx } }\) 
\(s = [ (y/3)\sqrt{10} ]_{1}^{8} \) \( (8/3)\sqrt{10}  (1/3)\sqrt{10} = 3\sqrt{10} \) 
Note  This is one of the very few times that we can evaluate the integral in both the x and ydirections. Most of the time, we will get an integral in one of the directions that cannot be evaluated.
Final Answer 

\(s = 3\sqrt{10}\) 
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\(\displaystyle{ y = \frac{1}{3}(x^2+2)^{3/2} }\); \( 0 \leq x \leq 1 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ y = \frac{1}{3}(x^2+2)^{3/2} }\) for \( 0 \leq x \leq 1 \). Give your answer in exact form.
Solution 

video by PatrickJMT 

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\(\displaystyle{ y = \frac{x^2}{2}  \frac{\ln(x)}{4} }\); \( 2 \leq x \leq 4 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ y = \frac{x^2}{2}  \frac{\ln(x)}{4} }\) for \( 2 \leq x \leq 4 \). Give your answer in exact form.
Solution 

video by PatrickJMT 

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\(\displaystyle{y = \frac{e^x + e^{x}}{2} }\); \( 0 \leq x \leq \ln 2 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{y = \frac{e^x + e^{x}}{2} }\) for \( 0 \leq x \leq \ln 2 \). Give your answer in exact form.
Final Answer 

3/4
Problem Statement 

Calculate the length of the arc \(\displaystyle{y = \frac{e^x + e^{x}}{2} }\) for \( 0 \leq x \leq \ln 2 \). Give your answer in exact form.
Solution 

Since the curve is given in the form \(y=f(x)\), we will use the equation \(\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2} ~dx}}\) to calculate the arc length.
\(\displaystyle{y = \frac{e^x + e^{x}}{2} \to y' = \frac{e^x  e^{x}}{2} }\) 
\(\displaystyle{s = \int_{0}^{\ln2}{\sqrt{1 + [y']^2} ~dx} }\) 
\(\displaystyle{s = \int_{0}^{\ln2}{\sqrt{1 + \left[ \frac{e^x  e^{x}}{2} \right]^2 } ~dx} }\) 
\(\displaystyle{s = \int_{0}^{\ln2}{\sqrt{\frac{4}{4} + \left[ \frac{(e^x  e^{x})^2}{4} \right] } ~dx} }\) 
\(\displaystyle{s = \frac{1}{2} \int_{0}^{\ln2}{\sqrt{ 4 + e^{2x}  2 + e^{2x} } ~dx } }\) 
\(\displaystyle{s = \frac{1}{2} \int_{0}^{\ln2}{\sqrt{ e^{2x} + 2 + e^{2x} } ~dx } }\) 
\(\displaystyle{s = \frac{1}{2} \int_{0}^{\ln2}{\sqrt{ [ e^x + e^{x} ]^2 } ~dx } }\) 
\(\displaystyle{s = \frac{1}{2} \int_{0}^{\ln2}{e^x + e^{x} ~dx } }\) 
\(\displaystyle{s = \frac{1}{2} [ e^x  e^{x} ]_0^{\ln 2} }\) 
\(\displaystyle{s = \frac{1}{2} [ e^{\ln 2}  e^{\ln 2}  1 + 1 ] }\) 
\(\displaystyle{s = \frac{1}{2} [ 2  1/2 ] }\) 
\(\displaystyle{s = \frac{1}{2} [ 3/2 ] = 3/4 }\) 
Final Answer 

3/4 
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\( y = x^{3/2} \); \( 0 \leq x \leq 4 \)
Problem Statement 

Calculate the length of the arc \( y = x^{3/2} \) for \( 0 \leq x \leq 4 \). Give your answer in exact form.
Final Answer 

\((8/27)(10^{3/2}1) \)
Problem Statement 

Calculate the length of the arc \( y = x^{3/2} \) for \( 0 \leq x \leq 4 \). Give your answer in exact form.
Solution 

For a written out solution to this problem, see this page.
video by MIT OCW 

Final Answer 

\((8/27)(10^{3/2}1) \) 
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\(y = \ln(\cos x) \); \( 0 \leq x \leq \pi/3 \)
Problem Statement 

Calculate the length of the arc \(y = \ln(\cos x) \) for \( 0 \leq x \leq \pi/3 \). Give your answer in exact form.
Final Answer 

\( \ln(2 + \sqrt{3}) \)
Problem Statement 

Calculate the length of the arc \(y = \ln(\cos x) \) for \( 0 \leq x \leq \pi/3 \). Give your answer in exact form.
Solution 

Since the curve is given in the form \(y=f(x)\), we will use the equation \(\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2} ~dx}}\) to calculate the arc length.
\(\displaystyle{y = \ln(\cos x) \to }\) \(\displaystyle{ y' = \frac{1}{\cos x} (\sin x) = \frac{\sin x}{\cos x} }\) 
\(\displaystyle{s = \int_{0}^{\pi/3}{\sqrt{1 + \left[ \frac{\sin x}{\cos x} \right]^2 } ~dx } }\) 
\(\displaystyle{s = \int_{0}^{\pi/3}{\sqrt{\frac{\cos^2 x}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x} } ~dx } }\) 
\(\displaystyle{s = \int_{0}^{\pi/3}{\sqrt{\frac{\cos^2 x + \sin^2 x}{\cos^2 x} } ~dx } }\) 
\(\displaystyle{s = \int_{0}^{\pi/3}{\frac{1}{\cos x} ~dx } }\) 
\(\displaystyle{s = \int_{0}^{\pi/3}{\sec x ~dx } }\) 
\(\ln \sec x + \tan x _{0}^{\pi/3} \) 
\(\ln \sec(\pi/3) + \tan(\pi/3)   \ln \sec(0) + \tan(0)  \) 
\(\ln 2 + \sqrt{3} \) 
Final Answer 

\( \ln(2 + \sqrt{3}) \) 
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\(\displaystyle{ y = \frac{3}{2} x^{2/3} }\); \( 0 \leq x \leq 1 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ y = \frac{3}{2} x^{2/3} }\) for \( 0 \leq x \leq 1 \). Give your answer in exact form.
Final Answer 

\(2^{3/2}  1 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ y = \frac{3}{2} x^{2/3} }\) for \( 0 \leq x \leq 1 \). Give your answer in exact form.
Solution 

built with GeoGebra 

We will use the integral \(\displaystyle{s = \int_{a}^{b}{\sqrt{1 + [f'(x)]^2} ~dx} }\) where \(f(x)=(3/2)x^{2/3}\)
\(f'(x)=(3/2)(2/3)x^{1/3} = x^{1/3} \) 
\(\displaystyle{s = \int_{0}^{1}{\sqrt{1 + [x^{1/3}]^2} ~dx} }\) 
\(\displaystyle{s = \int_{0}^{1}{\sqrt{1 + \frac{1}{x^{2/3}}} ~dx} }\) 
\(\displaystyle{s = \int_{0}^{1}{\sqrt{ \frac{x^{2/3}+1}{x^{2/3}} } ~dx} }\) 
\(\displaystyle{s = \int_{0}^{1}{\frac{\sqrt{x^{2/3}+1}}{x^{1/3}} ~dx } }\) 
Use substitution with \(u = x^{2/3} + 1 \to du = (2/3)x^{1/3} ~dx \) 
Convert the limits of integration. 
\(\displaystyle{s = \int_{1}^{2}{(3/2) u^{1/2} ~du } }\) 
\(\displaystyle{s = \left[ (3/2)\frac{u^{3/2}}{3/2} \right]_1^2 }\) 
\(\displaystyle{2^{3/2}  1^{3/2} }\) 
Final Answer 

\(2^{3/2}  1 \) 
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\(\displaystyle{y = 3 + \frac{4}{5}x^{5/4} }\); \( 0 \leq x \leq 1 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{y = 3 + \frac{4}{5}x^{5/4} }\) for \( 0 \leq x \leq 1 \). Give your answer in exact form.
Final Answer 

\(\displaystyle{\frac{8}{15}(1+\sqrt{2}) }\)
Problem Statement 

Calculate the length of the arc \(\displaystyle{y = 3 + \frac{4}{5}x^{5/4} }\) for \( 0 \leq x \leq 1 \). Give your answer in exact form.
Solution 

\(\displaystyle{y = 3 + \frac{4}{5}x^{5/4} \to y' = x^{1/4} }\) 
\(\displaystyle{\int_0^1{\sqrt{1+x^{1/2}} ~dx } }\) 
\(u = 1+x^{1/2} \to du = (1/2)x^{1/2} ~dx \) 
\(dx = 2x^{1/2} ~du \) 
\(\displaystyle{\int_{x=0}^{x=1}{u^{1/2} (2x^{1/2} ~du) } }\) 
Before we can integrate, we need to do two things. First, we need to convert the \(x^{1/2}\) term in the integrand to a term involving u. Second, we need to convert the limits of integration in terms of u. 
\(\displaystyle{\int_1^2{2 u^{1/2} (u1) ~ du } }\) 
\(\displaystyle{2 \int_1^2{u^{3/2}  u^{1/2} ~ du } }\) 
\(\displaystyle{2 \left[ \frac{u^{5/2}}{5/2}  \frac{u^{3/2}}{3/2} \right]_1^2 }\) 
Normally, we would evaluate with the limits of integration at this point. But the fractions look like they could get messy. So let's simplify those first. 
\(\displaystyle{2 \left[ \frac{2}{5}u^{5/2}  \frac{2}{3}u^{3/2} \right]_1^2 }\) 
\(\displaystyle{2 \left[ \frac{6}{15}u^{5/2}  \frac{10}{15}u^{3/2} \right]_1^2 }\) 
\(\displaystyle{\frac{4}{15} \left[ 3u^{5/2}  5u^{3/2} \right]_1^2 }\) 
\(\displaystyle{\frac{4}{15} [ ( 3(2)^{5/2}  5(2)^{3/2} )  ( 3  5 ) ] }\) 
\(\displaystyle{\frac{4}{15} [ 12\sqrt{2}  10\sqrt{2} + 2 ] }\) 
\(\displaystyle{\frac{8}{15} [ \sqrt{2} + 1 ] }\) 
Final Answer 

\(\displaystyle{\frac{8}{15}(1+\sqrt{2}) }\) 
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\(\displaystyle{y = \frac{x^4}{4}+\frac{1}{8x^2} }\); \( 2 \leq x \leq 4 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{y = \frac{x^4}{4}+\frac{1}{8x^2} }\) for \( 2 \leq x \leq 4 \). Give your answer in exact form.
Final Answer 

\(\displaystyle{\frac{7683}{128} }\)
Problem Statement 

Calculate the length of the arc \(\displaystyle{y = \frac{x^4}{4}+\frac{1}{8x^2} }\) for \( 2 \leq x \leq 4 \). Give your answer in exact form.
Solution 

built with GeoGebra 

\(\displaystyle{\int_2^4{\sqrt{1+[y']^2} ~dx } }\) 
\(\displaystyle{y = \frac{x^4}{4}+\frac{1}{8x^2} \to y' = x^3\frac{x^{3}}{4} }\) 
\(\displaystyle{\int_2^4{\sqrt{ 1 + \left[ x^3  \frac{1}{4x^3} \right]^2 } ~dx } }\) 
\(\displaystyle{\int_2^4{\sqrt{ 1 + \left[ \frac{4x^61}{4x^3} \right]^2 } ~dx } }\) 
\(\displaystyle{\int_2^4{\sqrt{ \frac{16x^6}{16x^6} + \frac{(4x^61)^2}{16x^6} } ~dx } }\) 
\(\displaystyle{\int_2^4{\sqrt{ \frac{16x^6 +16x^{12}8x^6+1}{16x^6} } ~dx } }\) 
\(\displaystyle{\int_2^4{\sqrt{ \frac{16x^{12}+8x^6+1}{16x^6} } ~dx } }\) 
\(\displaystyle{\int_2^4{\sqrt{ \frac{(4x^6+1)^2}{(4x^3)^2} } ~dx } }\) 
\(\displaystyle{\int_2^4{\frac{4x^6+1}{4x^3} ~dx } }\) 
\(\displaystyle{\int_2^4{\frac{4x^6}{4x^3} + \frac{1}{4x^3} ~dx } }\) 
\(\displaystyle{\int_2^4{x^3 + \frac{1}{4}x^{3} ~dx } }\) 
\(\displaystyle{\left[ \frac{x^4}{4} + \frac{1}{4} \frac{x^{2}}{2} \right]_2^4 }\) 
\(\displaystyle{\frac{1}{4} \left[ 4^4  2^4  \frac{( 4^{2}  2^{2} )}{2} \right] }\) 
\(\displaystyle{\frac{1}{4} \left[ 256  16  \frac{1}{32} + \frac{1}{8} \right] }\) 
\(\displaystyle{\frac{1}{4} \left[ 240 + \frac{3}{32} \right] }\) 
\(\displaystyle{\frac{1}{128} \left[ 7680 + 3 \right] }\) 
\(\displaystyle{\frac{7683}{128} }\) 
Final Answer 

\(\displaystyle{\frac{7683}{128} }\) 
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\(\displaystyle{y = \frac{x^3}{6} + \frac{1}{2x} }\); \( 1 \leq x \leq 3 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{y = \frac{x^3}{6} + \frac{1}{2x} }\) for \( 1 \leq x \leq 3 \). Give your answer in exact form.
Final Answer 

\(\displaystyle{\frac{14}{3} }\)
Problem Statement 

Calculate the length of the arc \(\displaystyle{y = \frac{x^3}{6} + \frac{1}{2x} }\) for \( 1 \leq x \leq 3 \). Give your answer in exact form.
Solution 

\(\displaystyle{y = \frac{x^3}{6} + \frac{1}{2x} \to y' = \frac{x^2}{2}  \frac{1}{2x^2} = \frac{x^41}{2x^2} }\) 
\(\displaystyle{\int_1^3{\sqrt{1 + \left[ \frac{x^41}{2x^2} \right]^2 } ~dx } }\) 
\(\displaystyle{\int_1^3{\sqrt{\frac{4x^4 + (x^41)^2}{4x^4} } ~dx } }\) 
\(\displaystyle{\int_1^3{\sqrt{\frac{4x^4 + x^8  2x^4+1}{4x^4} } ~dx } }\) 
\(\displaystyle{\int_1^3{\sqrt{\frac{x^8 + 2x^4 + 1}{4x^4} } ~dx } }\) 
\(\displaystyle{\int_1^3{\sqrt{\frac{(x^4+1)^2}{(2x^2)^2} } ~dx } }\) 
\(\displaystyle{\int_1^3{\frac{x^4+1}{2x^2} ~dx } }\) 
\(\displaystyle{\frac{1}{2} \int_1^3{x^2 + x^{2} ~dx } }\) 
\(\displaystyle{\frac{1}{2} \left[\frac{x^3}{3} + \frac{x^{1}}{1} \right]_1^3 }\) 
\(\displaystyle{\frac{1}{2} \left[\frac{x^3}{3}  \frac{1}{x} \right]_1^3 }\) 
\(\displaystyle{\frac{1}{2} \left[\frac{3^3}{3}  \frac{1}{3}  \frac{1}{3} + 1 \right] }\) 
\(\displaystyle{\frac{1}{2} \left[10  \frac{2}{3} \right] }\) 
\(\displaystyle{\frac{1}{2} \left[\frac{30}{3}  \frac{2}{3} \right] }\) 
\(\displaystyle{\frac{1}{2} \left[\frac{28}{3} \right] = \frac{14}{3} }\) 
Final Answer 

\(\displaystyle{\frac{14}{3} }\) 
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\(\displaystyle{ y = \frac{2}{3}(x^2+1)^{3/2} }\); \( 0 \leq x \leq 2 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ y = \frac{2}{3}(x^2+1)^{3/2} }\) for \( 0 \leq x \leq 2 \). Give your answer in exact form.
Solution 

For the written out solution to this problem by a different instructor, see this page.
video by Krista King Math 

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\(\displaystyle{ f(x) = \frac{1}{3}x^{3/2} }\); \( [0,4] \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ f(x) = \frac{1}{3}x^{3/2} }\) for \( [0,4] \). Give your answer in exact form.
Solution 

video by MIP4U 

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\(\displaystyle{ f(x) = \frac{x^2}{8}  \ln(x) }\); \( [1,e] \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ f(x) = \frac{x^2}{8}  \ln(x) }\) for \( [1,e] \). Give your answer in exact form.
Solution 

video by MIP4U 

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\(\displaystyle{ x = \frac{1}{3} \sqrt{y}(y3) }\); \( 1 \leq y \leq 9 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ x = \frac{1}{3} \sqrt{y}(y3) }\) for \( 1 \leq y \leq 9 \). Give your answer in exact form.
Final Answer 

\(32/3\)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ x = \frac{1}{3} \sqrt{y}(y3) }\) for \( 1 \leq y \leq 9 \). Give your answer in exact form.
Solution 

video by Krista King Math 

Final Answer 

\(32/3\) 
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\(\displaystyle{ x = \frac{y^4}{8} + \frac{1}{4y^2} }\); \( 1 \leq y \leq 2 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ x = \frac{y^4}{8} + \frac{1}{4y^2} }\) for \( 1 \leq y \leq 2 \). Give your answer in exact form.
Final Answer 

\( 33/16 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ x = \frac{y^4}{8} + \frac{1}{4y^2} }\) for \( 1 \leq y \leq 2 \). Give your answer in exact form.
Solution 

Since the curve is given in the form \(x=g(y)\), we will use the equation \(\displaystyle{ s = \int_{c}^{d}{ \sqrt{ 1 + [g'(y)]^2 } ~dy } }\) to calculate the arc length.
\(\displaystyle{ \frac{dx}{dy} = g'(y) = \frac{y^3}{2} + \frac{1}{2}(y^{3}) }\) 
\(\displaystyle{ s = \int_{1}^{2}{ \sqrt{ \frac{4}{4} + \frac{[ y^3  y^{3} ]^2}{4} } ~dy } }\) 
\(\displaystyle{ s = \frac{1}{2}\int_{1}^{2}{ \sqrt{ 4 + y^6  2 + y^{6} } ~dy } }\) 
\(\displaystyle{ s = \frac{1}{2}\int_{1}^{2}{ \sqrt{ y^6 + 2 + y^{6} } ~dy } }\) 
\(\displaystyle{ s = \frac{1}{2}\int_{1}^{2}{ \sqrt{ [ y^3 + y^{3} ]^2 } ~dy } }\) 
\(\displaystyle{ s = \frac{1}{2}\int_{1}^{2}{ y^3 + y^{3} ~dy } }\) 
\(\displaystyle{ s = \frac{1}{2}\left[ \frac{y^4}{4} + \frac{y^{2}}{2} \right]_1^2 }\) 
\(\displaystyle{ s = \frac{1}{2}\left[ \frac{2^4}{4}  \frac{2^{2}}{2}  \frac{1}{4} + \frac{1}{2} \right] }\) 
\(\displaystyle{ s = 2  \frac{1}{16} + \frac{1}{8} }\) 
\(\displaystyle{ s = \frac{32}{16}  \frac{1}{16} + \frac{2}{16} = \frac{33}{16} }\) 
Final Answer 

\( 33/16 \) 
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\(\displaystyle{x = \frac{y^5}{5} + \frac{1}{12y^3} }\); \( 1/2 \leq y \leq 1 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{x = \frac{y^5}{5} + \frac{1}{12y^3} }\) for \( 1/2 \leq y \leq 1 \). Give your answer in exact form.
Final Answer 

\( 373/480 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{x = \frac{y^5}{5} + \frac{1}{12y^3} }\) for \( 1/2 \leq y \leq 1 \). Give your answer in exact form.
Solution 

Since the curve is given in the form \(x=g(y)\), we will use the equation \(\displaystyle{s = \int_{c}^{d}{\sqrt{1 + [g'(y)]^2 } ~dy } }\) to calculate the arc length.
First, let\'s set up the integral.
\(\displaystyle{\frac{dx}{dy} = g'(y) = y^4 + \frac{1}{12}(3)y^{4} = y^4  \frac{y^{4}}{4} }\) 
\(\displaystyle{s = \int_{1/2}^{1}{\sqrt{1 + \left[ y^4  \frac{y^{4}}{4} \right]^2 } ~dy } }\) 
Now, let's simplify the term under the square root and integrate. 
\(\displaystyle{s = \int_{1/2}^{1}{\sqrt{1 + y^8  2\frac{1}{4} + \frac{y^{8}}{16} } ~dy } }\) 
\(\displaystyle{s = \int_{1/2}^{1}{\sqrt{y^8 + \frac{1}{2} + \frac{y^{8}}{16} } ~dy } }\) 
\(\displaystyle{s = \int_{1/2}^{1}{\sqrt{\frac{16y^8}{16} + \frac{8}{16} + \frac{y^{8}}{16} } ~dy } }\) 
\(\displaystyle{s = \frac{1}{4} \int_{1/2}^{1}{\sqrt{16y^8 + 8 + y^{8} } ~dy } }\) 
\(\displaystyle{s = \frac{1}{4} \int_{1/2}^{1}{\sqrt{(4y^4 + y^{4})^2 } ~dy } }\) 
\(\displaystyle{s = \frac{1}{4} \int_{1/2}^{1}{4y^4 + y^{4} ~dy } }\) 
\(\displaystyle{s = \frac{1}{4} \left[ \frac{4y^5}{5} + \frac{y^{3}}{3} \right]_{1/2}^1 }\) 
\(\displaystyle{s = \frac{1}{4} \left[ \frac{4}{5}  \frac{1}{3}  \frac{4(1/2)^5}{5} + \frac{(1/2)^{3}}{3} \right] }\) 
\(\displaystyle{s = \frac{1}{4} \left[ \frac{4}{5}  \frac{1}{3}  \frac{1}{40} + \frac{8}{3} \right] }\) 
\(\displaystyle{s = \frac{1}{4} \left[ \frac{4}{5}  \frac{1}{40} + \frac{7}{3} \right] }\) 
\(\displaystyle{s = \frac{1}{4} \left[ \frac{96}{120}  \frac{3}{120} + \frac{280}{120} \right] }\) 
\(\displaystyle{s = \frac{1}{4} \left[ \frac{373}{120} \right] }\) 
\(\displaystyle{s = \frac{373}{480} }\) 
Final Answer 

\( 373/480 \) 
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\(\displaystyle{ x = \ln y  \frac{y^2}{8} }\); \( 1 \leq y \leq 2 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ x = \ln y  \frac{y^2}{8} }\) for \( 1 \leq y \leq 2 \). Give your answer in exact form.
Final Answer 

\( 3/8 + \ln 2 \)
Problem Statement 

Calculate the length of the arc \(\displaystyle{ x = \ln y  \frac{y^2}{8} }\) for \( 1 \leq y \leq 2 \). Give your answer in exact form.
Solution 

Since the curve is given in the form \(x=g(y)\), we will use the equation \(\displaystyle{ s = \int_{c}^{d}{ \sqrt{ 1 + [g'(y)]^2 } ~dy } }\) to calculate the arc length.
First, let's set up the integral. 
\(\displaystyle{ \frac{dx}{dy} = g'(y) = \frac{1}{y}  \frac{1}{8}(2y) = \frac{1}{y}  \frac{y}{4} }\) 
\(\displaystyle{ s = \int_{1}^{2}{ \sqrt{ 1 + \left[ \frac{1}{y}  \frac{y}{4} \right]^2 } ~dy } }\) 
Now simplify the integrand. 
\(\displaystyle{s = \int_{1}^{2}{\sqrt{1 + \frac{1}{y^2} + \frac{y^2}{16}  \frac{1}{2} } ~dy } }\) 
\(\displaystyle{s = \int_{1}^{2}{\sqrt{\frac{1}{y^2} + \frac{y^2}{16} + \frac{1}{2} } ~dy } }\) 
\(\displaystyle{s = \int_{1}^{2}{\sqrt{\frac{16}{16y^2} + \frac{y^4}{16y^2} + \frac{8y^2}{16y^2} } ~dy } }\) 
\(\displaystyle{s = \int_{1}^{2}{\sqrt{\frac{16 + y^4 + 8y^2}{16y^2} } ~dy } }\) 
\(\displaystyle{s = \int_{1}^{2}{\sqrt{\frac{(y^2+4)^2}{16y^2} } ~dy } }\) 
\(\displaystyle{s = \int_{1}^{2}{\frac{(y^2+4)}{4y} ~dy } }\) 
\(\displaystyle{s = \frac{1}{4} \int_{1}^{2}{y + \frac{4}{y} ~dy } }\) 
Now we can integrate. 
\(\displaystyle{s = \frac{1}{4} \left[ \frac{y^2}{2} + 4\ln y \right]_1^2 }\) 
\(\displaystyle{s = \frac{1}{4} \left[ \frac{2^2}{2} + 4\ln 2  \frac{1^2}{2}  4\ln 1 \right] }\) 
\(\displaystyle{s = \frac{1}{2} + \ln 2  \frac{1}{8}  0 }\) 
\(\displaystyle{s = \frac{3}{8} + \ln 2 }\) 
Final Answer 

\( 3/8 + \ln 2 \) 
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You CAN Ace Calculus
related topics on other pages 

For arc length of parametric curves, see the parametrics calculus page. 
For the arc length of polar curves, see the polar calculus page. 
external links you may find helpful 
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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MultiVariable Calculus 

Differential Equations 

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Practice Instructions
Unless otherwise instructed, calculate 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 going through each solution will give you a broader point of view and help you better understand how to work these types problems.