# 6.3: Arc Length

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- Determine the length of a curve, \(x=g(y)\), between two points.
- Find the surface area of a solid of revolution.

In this section, we use definite integrals to find the arc length of a curve. We can think of **arc length** as the distance you would travel if you were walking along the path of the curve. Many real-world applications involve arc length. If a rocket is launched along a parabolic path, we might want to know how far the rocket travels. Or, if a curve on a map represents a road, we might want to know how far we have to drive to reach our destination.

We begin by calculating the arc length of curves defined as functions of \( x\), then we examine the same process for curves defined as functions of \( y\). (The process is identical, with the roles of \( x\) and \( y\) reversed.) The techniques we use to find arc length can be extended to find the surface area of a surface of revolution, and we close the section with an examination of this concept.

## Arc Length of the Curve y = f(x)

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 \( i=0,1,2,…,n\), let \( P={x_i}\) be a regular partition of \( [a,b]\). Then, for \( i=1,2,…,n\), 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. Figure \(\PageIndex{1}\) 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 \( Δx\). The change in vertical distance varies from interval to interval, though, so we use \( Δy_i=f(x_i)−f(x_{i−1})\) to represent the change in vertical distance over the interval \( [x_{i−1},x_i]\), as shown in Figure \(\PageIndex{2}\). Note that some (or all) \( Δy_i\) may be negative.

By the Pythagorean theorem, the length of the line segment is

\[ \sqrt{(Δx)^2+(Δy_i)^2}. \nonumber \]

We can also write this as

\[ Δx\sqrt{1+((Δy_i)/(Δx))^2}. \nonumber \]

Now, by the Mean Value Theorem, there is a point \( x^∗_i∈[x_{i−1},x_i]\) such that \( f′(x^∗_i)=(Δy_i)/(Δx)\). Then the length of the line segment is given by

\[ Δx\sqrt{1+[f′(x^∗_i)]^2}. \nonumber \]

Adding up the lengths of all the line segments, we get

\[\text{Arc Length} ≈\sum_{i=1}^n\sqrt{1+[f′(x^∗_i)]^2}Δx.\nonumber \]

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

\[\begin{align*} \text{Arc Length} &=\lim_{n→∞}\sum_{i=1}^n\sqrt{1+[f′(x^∗_i)]^2}Δx \\[4pt] &=∫^b_a\sqrt{1+[f′(x)]^2}dx.\end{align*}\]

We summarize these findings in the following theorem.

Let \( f(x)\) be a smooth function over the interval \([a,b]\). Then the arc length of the portion of the graph of \( f(x)\) from the point \( (a,f(a))\) to the point \( (b,f(b))\) is given by

\[\text{Arc Length}=∫^b_a\sqrt{1+[f′(x)]^2}\,dx. \nonumber \]

Note that we are integrating an expression involving \( f′(x)\), so we need to be sure \( f′(x)\) is integrable. This is why we require \( f(x)\) to be smooth. The following example shows how to apply the theorem.

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

**Solution**

We have \( f′(x)=3x^{1/2},\) so \( [f′(x)]^2=9x.\) Then, the arc length is

\[\begin{align*} \text{Arc Length} &=∫^b_a\sqrt{1+[f′(x)]^2}dx \nonumber \\[4pt] &= ∫^1_0\sqrt{1+9x}\,dx. \nonumber \end{align*}\]

Substitute \( u=1+9x.\) Then, \( du=9\,dx.\) When \( x=0\), then \( u=1\), and when \( x=1\), then \( u=10\). Thus,

\[ \begin{align*} \text{Arc Length} &=∫^1_0\sqrt{1+9x}\,dx \\[4pt] =\dfrac{1}{9}∫^1_0\sqrt{1+9x}\cdot 9\,dx \\[4pt] &= \dfrac{1}{9}∫^{10}_1\sqrt{u}\,du \\[4pt] &=\dfrac{1}{9}⋅\dfrac{2}{3}u^{3/2}\Big|^{10}_1 =\dfrac{2}{27}[10\sqrt{10}−1] \\[4pt] &≈2.268\,\text{units}. \end{align*}\]

Let \(f(x)=(4/3)x^{3/2}\). Calculate the arc length of the graph of \( f(x)\) over the interval \( [0,1]\). Round the answer to three decimal places.

**Hint**-
Use the process from the previous example. Don’t forget to change the limits of integration.

**Answer**-
\[ \dfrac{1}{6}(5\sqrt{5}−1)≈1.697 \,\text{units}\nonumber \]

Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in Introduction to Techniques of Integration. In some cases, we may have to use a computer or calculator to approximate the value of the integral.

Let \( f(x)=x^2\). Calculate the arc length of the graph of \( f(x)\) over the interval \( [1,3]\).

**Solution**

We have \( f′(x)=2x,\) so \( [f′(x)]^2=4x^2.\) Then the arc length is given by

\[\begin{align*} \text{Arc Length} &=∫^b_a\sqrt{1+[f′(x)]^2}\,dx \\[4pt] &=∫^3_1\sqrt{1+4x^2}\,dx. \end{align*}\]

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

\[ ∫^3_1\sqrt{1+4x^2}\,dx ≈ 8.26815 \,\text{units}. \nonumber \]

Let \( f(x)=\sin x\). Calculate the arc length of the graph of \( f(x)\) over the interval \( [0,π]\). Use a computer or calculator to approximate the value of the integral.

**Hint**-
Use the process from the previous example.

**Answer**-
\[ \text{Arc Length} ≈ 3.8202 \,\text{units}\nonumber \]

## Arc Length of the Curve x = g(y)

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 \(y\)-axis instead of the \(x\)-axis. Figure \(\PageIndex{3}\) shows a representative line segment.

Then the length of the line segment is

\[\sqrt{(Δy)^2+(Δx_i)^2}, \nonumber \]

which can also be written as

\[Δy\sqrt{1+\left(\dfrac{Δx_i}{Δy}\right)^2}. \nonumber \]

If we now follow the same development we did earlier, we get a formula for arc length of a function \(x=g(y)\).

Let \(g(y)\) be a smooth function over an interval \([c,d]\). Then, the arc length of the graph of \(g(y)\) from the point \((c,g(c))\) to the point \((d,g(d))\) is given by

\[\text{Arc Length}=∫^d_c\sqrt{1+[g′(y)]^2}\,dy. \nonumber \]

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

**Solution**

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

\[\begin{align*} \text{Arc Length} &=∫^d_c\sqrt{1+[g′(y)]^2}\,dy \\[4pt] &=∫^2_1\sqrt{1+81y^4}\,dy.\end{align*}\]

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

\[ ∫^2_1\sqrt{1+81y^4}dy≈21.0277 \,\text{units}.\nonumber \]

Let \(g(y)=1/y\). 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.

**Hint**-
Use the process from the previous example.

**Answer**-
\[\text{Arc Length} =3.15018 \,\text{units}\nonumber \]

## 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 \(x\)-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 \(i=0,1,2,…,n\), let \(P={x_i}\) be a regular partition of \([a,b]\). Then, for \(i=1,2,…,n,\) 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 \(x\)-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

\[\text{Lateral Surface Area } =πrs, \nonumber \]

where \(r\) is the radius of the base of the cone and \(s\) is the **slant height** (Figure \(\PageIndex{7}\)).

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 (Figure \(\PageIndex{8}\)).

The cross-sections of the small cone and the large cone are similar triangles, so we see that

\[ \dfrac{r_2}{r_1}=\dfrac{s−l}{s} \nonumber \]

Solving for \(s\), we get

\[\begin{align*} \dfrac{r_2}{r_1} &=\dfrac{s−l}{s} \\ r_2s &=r_1(s−l) \\ r_2s &=r_1s−r_1l \\ r_1l &=r_1s−r_2s \\ r_1l &=(r_1−r_2)s \\ \dfrac{r_1l}{r_1−r_2} &=s \end{align*}\]

Then the lateral surface area (SA) of the frustum is

\[\begin{align*} S &= \text{(Lateral SA of large cone)}− \text{(Lateral SA of small cone)} \\[4pt] &=πr_1s−πr_2(s−l) \\[4pt] &=πr_1\left(\dfrac{r_1l}{r_1−r_2}\right)−πr_2\left(\dfrac{r_1l}{r_1−r_2}-l\right) \\[4pt] &=\dfrac{πr^2_1l}{r^1−r^2}−\dfrac{πr_1r_2l}{r_1−r_2}+πr_2l \\[4pt] &=\dfrac{πr^2_1l}{r_1−r_2}−\dfrac{πr_1r2_l}{r_1−r_2}+\dfrac{πr_2l(r_1−r_2)}{r_1−r_2} \\[4pt] &=\dfrac{πr^2_1}{lr_1−r_2}−\dfrac{πr_1r_2l}{r_1−r_2} + \dfrac{πr_1r_2l}{r_1−r_2}−\dfrac{πr^2_2l}{r_1−r_3} \\[4pt] &=\dfrac{π(r^2_1−r^2_2)l}{r_1−r_2}=\dfrac{π(r_1−r+2)(r1+r2)l}{r_1−r_2} \\[4pt] &= π(r_1+r_2)l. \label{eq20} \end{align*} \]

Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the \(x\)-axis. A representative band is shown in the following figure.

Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have

\[\begin{align*} S &=π(r_1+r_2)l \\ &=π\big(f(x_{i−1})+f(x_i)\big)\sqrt{Δx^2+(Δy_i)^2} \\ &=π\big(f(x_{i−1})+f(x_i)\big)Δx\sqrt{1+\left(\dfrac{Δy_i}{Δx}\right)^2} \end{align*}\]

Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select \(x^∗_i∈\left[x_{i−1},x_i\right]\) such that \(f′(x^∗_i)=(Δy_i)/Δx.\) This gives us

\[S=π\big(f(x_{i−1})+f(x_i)\big)Δx\sqrt{1+\big(f′(x^∗_i)\big)^2} \nonumber \]

Furthermore, since \(f(x)\) is continuous, by the *Intermediate Value Theorem*, there is a point \(x^{**}_i∈\left[x_{i−1},x_i\right]\) such that \(f(x^{**}_i)=\tfrac{1}{2}\left[f(x_{i−1})+f(x_i)\right]\),

so we get

\[S=2πf(x^{**}_i)Δx\sqrt{1+\big(f′(x^∗_i)\big)^2}.\nonumber \]

Then the approximate surface area of the whole surface of revolution is given by

\[\text{Surface Area} ≈\sum_{i=1}^n 2πf(x^{**}_i)Δx\sqrt{1+\big(f′(x^∗_i)\big)^2}.\nonumber \]

This almost looks like a Riemann sum, except we have functions evaluated at two different points, \(x^∗_i\) and \(x^{**}_{i}\), over the interval \([x_{i−1},x_i]\). Although we do not examine the details here, it turns out that because \(f(x)\) is smooth, if we let \(n→∞\), the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both \(x^∗_i\) and \(x^{**}_i\) are in the interval \([x_{i−1},x_i]\), so it makes sense that as \(n→∞\), both \(x^∗_i\) and \(x^{**}_i\) approach \(x\) Those of you who are interested in the details should consult an advanced calculus text.

Taking the limit as \(n→∞,\) we get

\[ \begin{align*} \text{Surface Area} &=\lim_{n→∞}\sum_{i=1}^n 2πf(x^{**}_i)Δx\sqrt{1+\big(f′(x^∗_i)\big)^2} \\[4pt] &=∫^b_a\left(2πf(x)\sqrt{1+\big(f′(x)\big)^2}\right)\,dx \end{align*}\]

As with arc length, we can conduct a similar development for functions of \(y\) to get a formula for the surface area of surfaces of revolution about the \(y\)-axis. These findings are summarized in the following theorem.

Let \(f(x)\) be a nonnegative smooth function over the interval \([a,b]\). Then, the surface area of the surface of revolution formed by revolving the graph of \(f(x)\) around the \(x\)-axis is given by

\[\text{Surface Area}=∫^b_a\left(2πf(x)\sqrt{1+\big(f′(x)\big)^2}\right)dx \nonumber \]

Similarly, let \(g(y)\) be a nonnegative smooth function over the interval \([c,d]\). Then, the surface area of the surface of revolution formed by revolving the graph of \(g(y)\) around the \(y\)-axis is given by

\[\text{Surface Area}=∫^d_c\left(2πg(y)\sqrt{1+\big(g′(y)\big)^2}\right)dy \nonumber \]

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 \(x\)-axis. Round the answer to three decimal places.

**Solution**

The graph of \(f(x)\) and the surface of rotation are shown in Figure \(\PageIndex{10}\).

We have \(f(x)=\sqrt{x}\). Then, \(f′(x)=\frac{1}{2\sqrt{x}}\) and \(\big(f′(x)\big)^2=\frac{1}{4x}.\) Then,

\[\begin{align*} \text{Surface Area} &=∫^b_a\left(2πf(x)\sqrt{1+\big(f′(x)\big)^2}\right)dx \\[4pt] &=∫^4_1\left(2π\sqrt{x}\sqrt{1+\dfrac{1}{4x}}\right)\,dx \\[4pt] &=∫^4_1\left(2π\sqrt{x+\frac{1}{4}}\right)\,dx. \end{align*}\]

Let \(u=x+1/4.\) Then, \(du=dx\). When \(x=1, \, u=5/4\), and when \(x=4, \, u=17/4.\) This gives us

\[\begin{align*} ∫^1_0\left(2π\sqrt{x+\dfrac{1}{4}}\right)\,dx &= ∫^{17/4}_{5/4}2π\sqrt{u}\,du \\[4pt] &= 2π\left[\dfrac{2}{3}u^{3/2}\right]∣^{17/4}_{5/4} \\[4pt] &=\dfrac{π}{6}[17\sqrt{17}−5\sqrt{5}]≈30.846 \,\text{units}^2 \end{align*}\]

Let \( f(x)=\sqrt{1−x}\) over the interval \( [0,1/2]\). Find the surface area of the surface generated by revolving the graph of \( f(x)\) around the \(x\)-axis. Round the answer to three decimal places.

**Hint**-
Use the process from the previous example.

**Answer**-
\[ \dfrac{π}{6}\left(5\sqrt{5}−3\sqrt{3}\right)≈3.133 \,\text{units}^2\nonumber \]

Let \( f(x)=y=\sqrt[3]{3x}\). Consider the portion of the curve where \( 0≤y≤2\). Find the surface area of the surface generated by revolving the graph of \( f(x)\) around the \( y\)-axis.

**Solution**

Notice that we are revolving the curve around the \( y\)-axis, and the interval is in terms of \( y\), so we want to rewrite the function as a function of \( y\). We get \( x=g(y)=(1/3)y^3\). The graph of \( g(y)\) and the surface of rotation are shown in the following figure.

We have \( g(y)=(1/3)y^3\), so \( g′(y)=y^2\) and \( \big(g′(y)\big)^2=y^4\). Then

\[\begin{align*} \text{Surface Area} &=∫^d_c\left(2πg(y)\sqrt{1+\big(g′(y)\big)^2}\right)\,dy \\[4pt] &=∫^2_0\left(2π\left(\frac{1}{3}y^3\right)\sqrt{1+y^4}\right)\,dy \\[4pt] &=\frac{2π}{3}∫^2_0\left(y^3\sqrt{1+y^4}\right)\,dy. \end{align*}\]

Let \( u=y^4+1.\) Then \( du=4y^3\,dy\). When \( y=0, \, u=1\), and when \( y=2, \, u=17.\) Then

\[\begin{align*} \frac{2π}{3}∫^2_0\left(y^3\sqrt{1+y^4}\right)\,dy &=\frac{2π}{3}∫^{17}_1\frac{1}{4}\sqrt{u}\,du \\[4pt] &=\frac{π}{6}\left[\dfrac{2}{3}u^{3/2}\right]\Big|^{17}_1=\frac{π}{9}\left[(17)^{3/2}−1\right]≈24.118 \,\text{units}^2. \end{align*}\]

Let \( g(y)=\sqrt{9−y^2}\) over the interval \( y∈[0,2]\). Find the surface area of the surface generated by revolving the graph of \( g(y)\) around the \( y\)-axis.

**Hint**-
Use the process from the previous example.

**Answer**-
\( 12π \,\text{units}^2\)

## 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**

**Arc Length** \( \displaystyle =∫^b_a\sqrt{1+[f′(x)]^2}\,dx\)

**Arc Length of a Function of y**

**Arc Length **\( \displaystyle =∫^d_c\sqrt{1+[g′(y)]^2}\,dy\)

**Surface Area of a Function of x**

**Surface Area** \( \displaystyle =∫^b_a\left(2πf(x)\sqrt{1+\big(f′(x)\big)^2}\right)\,dx\)

## Glossary

**arc length**- the arc length of a curve can be thought of as the distance a person would travel along the path of the curve

**frustum**- a portion of a cone; a frustum is constructed by cutting the cone with a plane parallel to the base

**surface area**- the surface area of a solid is the total area of the outer layer of the 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