I'm working on arc length calculation and area of surface of revolution in calculus and I'm really quite stuck on the process of how to do this. Here is a particular problem that I'm struggling with:

Find the surface area of the surface of revolution generated by revolving the graph $$y=x^3; \qquad 0 \leq x \leq10$$ around the $x$-axis.

I've gone through the steps that I've learned to do (listed below) and the steps for the most part seem to make sense, however I keep ending up with incorrect answers. Please help! Below I listed my general process of approaching the problem.

$$\begin{align} &y = x^3 \\&y' = 3x^2 \\&(y')^2 = 9x^4\\&1 + \left(\dfrac{dy}{dx}\right)^2 = 1 + 9x^4\end{align}$$

Using the formula:

$$2\pi y \cdot\int\left(1 + \left(\dfrac{dy}{dx}\right)^2\right)^{1/2} dx$$

Here's how I set up the integral for the problem:

$$2\pi x^3\cdot \int_0^{10}(1+9x^4)^{1/2} dx.$$

This came out to $$2\pi x^3\cdot\left(\dfrac{6}{5}\right)\cdot(1+9(10^4))^{3/2}\cdot x^5$$ My final answer was $$2.035785969E16.$$

Please help me understand where I'm going wrong!

  • 1
    $\begingroup$ $X^3$ should be placed inside the integration not outside $\endgroup$ May 6, 2017 at 13:29

1 Answer 1


After carefully reviewing your problem, I've identified a small misunderstanding that might have led to the confusion. It seems like you've taken $y$ out of the integral formula, which is a common misconception. In this context, $y$ actually represents the function itself or in another form $f(x)$, not a constant. I understand how this can happen, as these concepts can sometimes be a bit tricky to grasp.

Let's work through it together to make sure we're all on the same page and can learn from this experience. The correct approach involves keeping $y$ within the integral, as it's a function that varies with $x$. This might lead to a clearer solution. Feel free to ask further questions or share your thoughts so we can help you reach the correct solution.

Arc Length of a Curve

The arc length is the distance between two points that move along a section of a Cartesian coordinate curve. It can be approximated by a straight line, whose length can be defined by the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ on the Cartesian coordinate plane. $d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\tag*{}$ A rectifiable curve refers to a curve that has a finite arc length. A sufficient condition for a graph of a function to be rectifiable between $(a, f(a))$ and $(b, f(b))$ is that $f'(x)$ is continuous on the closed interval, $a \le x \le b$. This means that the function is continuously differentiable on that interval, and the graph on the interval $[a, b]$ is a smooth curve. Assume that the function is considered to be $y = f(x)$ that's continuously differentiable on the closed interval $[a, b]$, then the approximation to the graph of $f(x)$ by many line segments whose endpoints are determined by the partition, $$a=x_0<x_1<x_2<\ldots<x_n=b$$ as shown in Figure 1.

Approximation for the arc length of a curve

Then by letting $\Delta x_k=x_k-x_{k-1}$ and $\Delta y_k=y_k-y_{k-1}$, thus the approximate the length of the graph by $$\begin{aligned} \displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\implies\ell &= \sum_{k=1}^{n} \sqrt{(x_k-x_{k-1})^2+(y_k-y_{k-1})^2} \\ &=\sum_{k=1}^{n}\sqrt{\Delta {x_k}^2+\Delta{y_k}^2} \\ &=\sum_{k=1}^{n} \sqrt{\Delta {x_k}^2+\left( \frac{\Delta y_k}{\Delta x_k} \right)^2\Delta {x_k}^2} \\ &=\sum_{k=1}^n\sqrt{1+\left(\frac{\Delta y_k}{\Delta x_k} \right)^2} \,\Delta x_k \end{aligned}$$

Since many of the subintervals that were approximated were very big, which will make the infinite line segments on the endpoints, this would better appear as we take the limit of the length of the graph as $n$ approaches infinity. $$\ell =\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\sqrt{1+\left( \frac{\Delta y_k}{\Delta x_k} \right)^2} \,\Delta x_k$$

Since $f'(x)$ exists for each $x \in (x_{k-1},x_k)$, then the Mean Value Theorem guarantees the existence of $c_k \in (x_{k-1}, x_k)$ such that, $$\begin{aligned}\displaystyle f(x_k)-f(x_{k-1}) &=f'(c_k)(x_k-x_{k-1})\\\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}} &=f'(c_k) \\ \frac{\Delta y_k}{\Delta x_k} &=f'(c_k) \end{aligned}$$

also because $f'(x)$ is continuous on the closed interval $[a,b]$, which follows that $\sqrt{1+\left[f'(x)\right]^2}$ is also continuous and therefore integrable on $[a, b]$, which implies that,

$$\ell=\displaystyle \lim_{n\to\infty}\sum_{k=1}^n \sqrt{1+[f'(c_k)]^2} \,\Delta x_k=\int\limits_a^b\sqrt{1+[f'(x)]^2} \,\mathrm dx$$


Theorem 1: Arc Length

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 $(a,f(a))\to (b,f(b))$ is given by $$\ell = \displaystyle\int\limits^{b}_{a}\sqrt{1+\left[ f'\left( x\right) \right] ^{2}}\,\mathrm dx \iff \int\limits^{b}_{a}\sqrt{1+\left[ \dfrac{\mathrm dy}{\mathrm dx} \right] ^{2}}\,\mathrm dx$$ Similar to this if the arc length goes to $x=g(y)$ $$\ell = \displaystyle\int\limits^{d}_{c}\sqrt{1+\left[ g'\left( y\right) \right] ^{2}}\,\mathrm dy \iff \int\limits^{d}_{c}\sqrt{1+\left[ \dfrac{\mathrm dy}{\mathrm dx} \right] ^{2}}\,\mathrm dy$$

Example: $\text{1. Find the arc length of the curve $y=\displaystyle \frac{2x^{\frac32}}{3}$ from $\left( 0, \dfrac{2}{3} \right)$ to $\left(3, \dfrac{4\sqrt2}{3} \right)$}$ Solution: The first thing we need to do is take the derivative of the function. $\left.\begin{aligned} y= f(x)=\frac{2x^{\frac32}}{3}\implies\dfrac{\mathrm{d}}{\mathrm{d}x}\left[\frac{2x^{\frac32}}{3} \right] &=\frac{2}{3}\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ x^{\frac{3}{2}} \right] \\ &=\frac23\cdot\frac32x^{\frac12}\\ &= x^{\frac12} \equiv \sqrt x \end{aligned}\right\}\implies\dfrac{\mathrm dy}{\mathrm dx} =f'(x)=x^{\frac12}\tag*{}$ Then complete the definite integral. $\begin{aligned} \ell &=\displaystyle \int\limits^{b}_{a}\sqrt{1+\left[ \dfrac{\mathrm dy}{\mathrm dx} \right] ^{2}}\,\mathrm dx \\ &=\int\limits_0^3\sqrt{1+\left( \sqrt{x} \right)^2} \,\mathrm dx \\ &= \int\limits_0^3\sqrt{1+x} \,\mathrm dx \\ &=\int\limits_1^4\sqrt u \,\mathrm du \implies u=1+x\quad \mathrm du =\mathrm dx \in [0,3]\to[1,4] \\ &=\left.\frac{2u^{\frac32}}{3}\right|^4_1 \\ &=\frac{2\left( 4 \right)^{\frac{3}{2}}}{3}-\frac{2\left( 1 \right)^{\frac{3}{2}}}{3} \\ &=\frac{16}{3}-\frac{2}{3} \\ &=\frac{14}{3} \\ &=4\frac23 \end{aligned}\tag*{}$ $\text{Find the arc length of the curve $x^2=(y-4)^3$ that corresponds to the $y-$interval [32, 48].}$ Solve the equation and take the derivatives $\begin{array}{rcl} x^2&= &(y-4)^3 \\ x&=& \pm\sqrt{(y-4)^3} &\implies &\pm(y-4)^{\frac{3}{2}} \end{array}\tag*{}$ Since the equation provide two possible answers, we can choose the positive values of $x$ produces, $\dfrac{\mathrm{d}}{\mathrm{d}y}\left[ (y-4)^{\frac{3}{2}} \right]\tag*{}$ Use the Generalized Power Rule of the Derivatives *If *$y=\left[u\left(x\right)\right]^n$, *where $u$ is a differentiable function of $x$ and *$n\in \mathbb R$, then $\begin{array}{|c|} \hline \text{Generalized Power Rule} \\ \hline \dfrac{\mathrm dy}{\mathrm dx}=n\left[ u\left( x\right) \right] ^{n-1}\dfrac{\mathrm du}{\mathrm dx} \qquad \text{or} \qquad \dfrac{\mathrm{d}}{\mathrm{d}x}\left[ u^n \right]=nu^{n-1}u'\\ \hline \end{array}\tag*{}$ Hence $\begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}y}\left[ (y-4)^{\frac{3}{2}} \right] &=\frac{3}{2}(y-4)^{\frac{3}{2}-1}\dfrac{\mathrm{d}}{\mathrm{d}y}[y-4] \\ &=\frac32(y-4)^{\frac{1}{2}}(1) \\ &= \frac32(y-4)^{\frac{1}{2}}\end{aligned}\tag*{}$ Therefore $\begin{aligned} \ell =\int\limits^{d}_{c}\sqrt{1+\left[ \dfrac{\mathrm dy}{\mathrm dx} \right] ^{2}}\,\mathrm dy &=\int\limits_{32}^{48}\sqrt{1+\left( \frac{3}{2}(y-4)^{\frac{1}{2}} \right)^2} \,\mathrm dy \\ &=\int\limits_{32}^{48}\sqrt{1+\left( \frac{9}{4}(y-4)^{^\frac{1}{2}} \right)}\,\mathrm dy \\ &=\int\limits_{32}^{48}\sqrt{\frac{9}{4}y-8}\,\mathrm dy \end{aligned}\tag*{}$ Let, $u=\frac94y-8 \implies \mathrm du=\frac49\,\mathrm dy$. Then, $\begin{aligned} \displaystyle\int\limits_{32}^{48}\sqrt{\frac{9}{4}y-8}\,\mathrm dy &=\frac{4}{9}\int\limits_{64}^{100} \sqrt u\,\mathrm du \\ &=\frac49\left[\frac{2u^{^{\frac{3}{2}}}}{3} \right]^{100}_{64} \\ &=\frac49\left( \frac{2(100)^{\frac{3}{2}}}{3} -\frac{2\left( 64 \right)^{\frac{3}{2}}}{3}\right) \\ &=\frac{4}{9}\left( \frac{2000}{3}-\frac{1024}{3} \right) \\ &=\frac49\left( \frac{976}{3} \right) \\ &= \frac{3904}{27} \\ &=144\frac{16}{27} \end{aligned}$

Surface of Revolution

Lateral surface area of frustum of cone

The surface of revolution is formed from the derivation of the formula for the lateral surface area of the frustum (Latin for 'morsel') of a right circular cone. So, let's consider the cone that is circular on the line segment as shown in Figure 2, where $L$ is the length of the line segment and $r_1$ and $r_2$ are two radiuses that represent the left end and right end segments, respectively. When this line segment is revolved about its axis of revolution, it forms the form of a right circular cone, with $$S=2\pi rL \tag*{}$$ where $r=\frac12(r_1+r_2)$ is an average radius of frustum.

Now let's see the graph and when it's revolving as $360^{\circ}$ as shown in Figure 3.

The changes in shape occur when the 2D graph line form is revolved <span class=$360^{\circ}$ on the horizontal axis of revolution. " />

Now let us consider a function that has a continuous derivative on the interval $[a, b]$. The graph of the function that revolves around the axis of revolution, which we take about the horizontal axis to form a surface of revolution, is shown in Figure 3. So assuming the $\Delta$ is a partition of the closed interval $[a, b]$, with many subintervals of width $\Delta x_k$, Then for the line segment of length, we can take the previous form that was shown in our discussion of the arc length and generates a frustum of a cone. Let $r_k$ be the average radius of this frustum then by the Intermediate Value Theorem, a point $d_k$ exists in the $k$th such that we obtain the lateral surface area $\Delta S_k$ of the frustum that is expressed by this following.

$$\begin{aligned} \Delta L_k=\sqrt{\Delta{x_k}^2+\Delta {y_{k}}^2} \implies r_k=f(d_k) \implies S_k&=2\pi rL\\ \Delta S_k&=2\pi r_k\Delta L_k \\ &=2\pi f(d_k)\sqrt{\Delta{x_k}^2+\Delta {y_{k}}^2} \\ &=2\pi f(d_k)\sqrt{1+\left( \frac{\Delta y_k}{\Delta x_k} \right)^2} \,\Delta x_k \end{aligned}$$ By using Mean Value Theorem, a point $c_k \;\text{exists}\;\in(x_{k-1},x_k)$ such that $$\begin{aligned}\displaystyle f(x_k)-f(x_{k-1}) &=f'(c_k)(x_k-x_{k-1})\\\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}} &=f'(c_k) \\ \frac{\Delta y_k}{\Delta x_k} &=f'(c_k) \end{aligned}$$

So, $\Delta S_k =2\pi f(d_k)\sqrt{1+[f’(c_k)]^2}\,\Delta x_k$ and the total surface area can be approximated by Riemann sum as it becomes better with the limits of Riemann sum as many subintervals approach infinite ($n\to \infty$). Such that, $$\begin{aligned}\displaystyle S\approx \lim_{n\to\infty}\sum_{k=1}^{n}\Delta S_k \implies S &\approx\lim_{n\to \infty} \sum_{k=1}^n 2\pi f(d_k)\sqrt{1+[f'(c_k)]^2} \,\Delta x_k \\ &=2\pi\left( \lim_{n\to \infty}\sum_{k=1}^{n} f(d_k)\sqrt{1+[f'(c_k)]^2} \,\Delta x_k \right) \\ &=2\pi \int\limits_{a}^b f(x)\sqrt{1+[f'(x)]^2}\,\mathrm dx \iff 2\pi \int \limits_a^b f(x)\sqrt{1+\left[ \dfrac{\mathrm{d}y}{\mathrm{d}x} \right]^2}\,\mathrm dx \end{aligned}$$

Hence, it creates the Theorem that defines the surface of revolution by the definite integral.

Theorem 2: Surface of Revolution Let $f(x)$ be a nonnegative smooth function over the interval $[a,b]$. Then, the surface area of the surface revolution formed by revolving the graph of $f(x)$ around the horizontal axis of the revolution is given by $$\displaystyle S= 2\pi \int\limits_{a}^b f(x)\sqrt{1+[f'(x)]^2}\,\mathrm dx \iff 2\pi \int \limits_a^b f(x)\sqrt{1+\left[ \dfrac{\mathrm{d}y}{\mathrm{d}x} \right]^2}\,\mathrm dx $$ Similarly, if $g(y)$ be a nonnegative smooth function over the interval $[c,d]$. Then, the surface area of the surface revolution formed by revolving the graph of $g(y)$ around the horizontal axis of the revolution is given by $$\displaystyle S= 2\pi \int\limits_{c}^d g(y)\sqrt{1+[g'(y)]^2}\,\mathrm dy \iff 2\pi \int \limits_c^d g(y)\sqrt{1+\left[ \dfrac{\mathrm{d}x}{\mathrm{d}y} \right]^2}\,\mathrm dy$$

Now let's see this progress. Find the surface area of the surface of the revolution generated by revolving the graph $y=x^3;\qquad 0\le x\le 10$.


Firstly we have to find the derivative as the formula shown to us. $$\begin{aligned} y&=x^3 \\ \dfrac{\mathrm{d}y}{\mathrm{d}x}&=\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ x^3 \right] \\ y'&=3x^2 \end{aligned}$$ Apply the powers of 2 and integrate it by using the Fundamental Theorem of Calculus, $\displaystyle\int\limits_a^b f(x)\,\mathrm dx= F(b)-F(a)$.

$$\begin{aligned} S=\displaystyle 2\pi \int \limits_a^b y\sqrt{1+\left( y' \right)^2}\,\mathrm dx &= 2\pi \int \limits_a^b f(x)\sqrt{1+\left[ \dfrac{\mathrm{d}y}{\mathrm{d}x} \right]^2}\,\mathrm dx \\ &=2\pi \int\limits_0^{10}x^3\sqrt{1+(3x^2)^2}\,\mathrm dx \\ &=2\pi\int\limits_0^{10}x^3\sqrt{1+9x^4}\,\mathrm dx \end{aligned}$$

Apply the integration by substitution:

Suppose that, $u=1+9x^4$. This will obtain our differentiation and its differential form for the definite integral.

$$\begin{aligned} \dfrac{\mathrm{d}}{\mathrm{d}x}[u]&=\dfrac{\mathrm{d}}{\mathrm{d}x}\left[ 1+9x^4 \right] \\ \dfrac{\mathrm{d}u}{\mathrm{d}x} &=36x^3\end{aligned} \qquad \begin{aligned} \mathrm du&=36x^3\,\mathrm dx \\ \mathrm dx&= \dfrac{1}{36x^3}\,\mathrm du \end{aligned}$$

Thus the limits of integration, $x=0\implies u=1$ and $x=10\implies u=90001$. Hence

$$\begin{aligned} S=\displaystyle 2\pi\int\limits_{0}^{10}x^3\sqrt{1+9x^4}\,\mathrm dx &=2\pi\int\limits_{1}^{90001}x^3\sqrt{u}\left( \frac{1}{36x^3} \right) \,\mathrm du \\ &=2\pi\left( \frac{1}{36} \right) \int\limits_{1}^{90001}\sqrt u\,\mathrm du \\ &=\frac{\pi}{18}\int\limits_1^{90001}u^{\frac12}\,\mathrm du \\ &=\frac{\pi}{18}\left[ \frac{2u^{\frac32}}{3} \right]^{90001}_{1}\\ &=\frac{\pi}{18}\left[ \frac{2(90001)^{\frac32}}{3}-\frac{2(1)^{\frac32}}{3} \right] \\ &=\frac{\pi}{18}\left[ \frac{2(90001)^{\frac32}-2}{3} \right] \\ &=\frac{\pi\left( 90001^{\frac32}-1 \right)}{27} \approx 3141644.\overline{8972} \end{aligned}$$

I hope this explanation and demonstration have provided you with a clearer understanding of how to approach the surface of the revolution problem. If you have any further questions or need additional assistance, please feel free to ask. Your feedback and comments are always welcome and appreciated as they help improve the quality of information shared here.


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