8.2: Surface Area

Another geometric question that arises naturally is: "What is the surface area of a volume?'' For example, what is the surface area of a sphere? More advanced techniques are required to approach this question in general, but we can compute the areas of some volumes generated by revolution.

As usual, the question is: how might we approximate the surface area? For a surface obtained by rotating a curve around an axis, we can take a polygonal approximation to the curve, as in the last section, and rotate it around the same axis. This gives a surface composed of many "truncated cones;'' a truncated cone is called a frustum of a cone. Figure \(\PageIndex{1}\) illustrates this approximation.

Figure \(\PageIndex{1}\): Approximating a surface (left) by portions of cones (right). You can download the Sage worksheetfor this plot and upload it to your own sage account.

So we need to be able to compute the area of a frustum of a cone. Since the frustum can be formed by removing a small cone from the top of a larger one, we can compute the desired area if we know the surface area of a cone. Suppose a right circular cone has base radius \(r\) and slant height \(h\). If we cut the cone from the vertex to the base circle and flatten it out, we obtain a sector of a circle with radius \(h\) and arc length \(2\pi r\), as in Figure \(\PageIndex{2}\). The angle at the center, in radians, is then \(\dfrac{2\pi r}{h}\), and the area of the cone is equal to the area of the sector of the circle. Let \(A\) be the area of the sector; since the area of the entire circle is \(\pi h^2\), we have

\[\dfrac{A}{\pi h^2} =\dfrac{\frac{2\pi r}{h}}{ 2\pi}  \implies  A = \pi r h  \nonumber \]

Figure \(\PageIndex{2}\): The area of a cone.

Now suppose we have a frustum of a cone with slant height \(h\) and radii \(r_0\) and \(r_1\), as in Figure \(\PageIndex{3}\). The area of the entire cone is \(\pi r_1(h_0+h)\), and the area of the small cone is \(\pi r_0 h_0\); thus, the area of the frustum is \(\pi r_1(h_0+h)-\pi r_0 h_0=\pi((r_1-r_0)h_0+r_1h)\). By similar triangles,

\[{h_0\over r_0}={h_0+h\over r_1} \implies (r_1-r_0)h_0=r_0h\nonumber \]

Substitution into the area gives

\[\pi((r_1-r_0)h_0+r_1h)=\pi(r_0h+r_1h)=\pi h(r_0+r_1)=2\pi \dfrac{r_0+r_1}{2} h = 2\pi r h \nonumber \]

The final form is particularly easy to remember, with \(r\) equal to the average of \(r_0\) and \(r_1\), as it is also the formula for the area of a cylinder. (Think of a cylinder of radius \(r\) and height \(h\) as the frustum of a cone of infinite height.)

Figure \(\PageIndex{3}\): The area of a frustum.

Now we are ready to approximate the area of a surface of revolution. On one subinterval, the situation is as shown in Figure \(\PageIndex{4}\). When the line joining two points on the curve is rotated around the \(x\)-axis, it forms a frustum of a cone. The area is

\[ 2\pi r h= 2\pi \dfrac{f(x_i)+f(x_{i+1}}{2} \sqrt{1+\left(f'(t_i)\right)^2}\,\Delta x \nonumber \]

Here \(\sqrt{1+\left(f'(t_i)\right)^2}\,\Delta x\) is the length of the line segment, as we found in the previous section. Assuming \(f\) is a continuous function, there must be some \(x_i^*\) in \( [x_i,x_{i+1}]\) such that \(\dfrac{f(x_i)+f(x_{i+1})}{2} = f(x_i^*)\), so the approximation for the surface area is

\[\sum_{i=0}^{n-1} 2\pi f(x_i^*)\sqrt{1+\left(f'(t_i)\right)^2}\,\Delta x \nonumber \]

This is not quite the sort of sum we have seen before, as it contains two different values in the interval \([x_i,x_{i+1}]\), namely \(x_i^*\) and \(t_i\). Nevertheless, using more advanced techniques than we have available here, it turns out that

\[\lim_{n\to\infty} \sum_{i=0}^{n-1} 2\pi f(x_i^*)\sqrt{1+\left(f'(t_i)\right)^2}\,\Delta x= \int_a^b 2\pi f(x)\sqrt{1+\left(f'(x)\right)^2}\,dx \nonumber \]

is the surface area we seek. (Roughly speaking, this is because while \(x_i^*\) and \(t_i\) are distinct values in \([x_i,x_{i+1}]\), they get closer and closer to each other as the length of the interval shrinks.)

Figure \(\PageIndex{4}\): One subinterval.

If the curve is rotated around the \(y\) axis, the formula is nearly identical, because the length of the line segment we use to approximate a portion of the curve doesn't change. Instead of the radius \(f(x_i^*)\), we use the new radius \(\bar x_i= (x_i+x_{i+1})/2\), and the surface area integral becomes

\[\int_a^b 2\pi x\sqrt{1+\left(f'(x)\right)^2}\,dx \nonumber \]

These results are summarized in the following Theorem.

Theorem \(\PageIndex{1}\)

If the curve \(y=f(x)\) for \(a\le x\le b\) is rotated around the \(x\)-axis, the surface area of the resulting object is

\[\int_a^b 2\pi f(x)\sqrt{1+\left(f'(x)\right)^2}\,dx \nonumber \]

If the curve \(y=f(x)\) for \(a\le x\le b\) is rotated around the \(y\)-axis, the surface area of the resulting object is

\[ \int_a^b 2\pi x\sqrt{1+\left(f'(x)\right)^2}\,dx \nonumber \]

Example \(\PageIndex{1}\)

Let \(f(x)\) be the straight line connecting \((0,3)\) to \((4,0)\). Find the surface area of the cone formed when \(y=f(x)\) is rotated around

a. the \(x\)-axis

b. the \(y\)-axis

Solution

First, we find the equation of the described line by find the slope \(m=\dfrac{0-3}{4-0}=-\dfrac{3}{4}\).

Using point-slope form of a line, we can easily find that the line is therefore \(y=-\dfrac{3}{4}x+3\).

Each of these surface areas of revolutions will require us to solve an integral containing \(\sqrt{1+\left(f'(x)\right)^2}\), so we evaluate this expression for \(f(x)=-\dfrac{3}{4}x+3\):

\[ f'(x)=-\dfrac{3}{4} \implies \sqrt{1+\left(f'(x)\right)^2}=\sqrt{1+\left(-\dfrac{3}{4}\right)^2}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4} \nonumber \]

(a) The surface area when this line is rotated around the \(x\)-axis is given by:

\[\begin{align*} \text{Surface Area} &=\int_a^b 2\pi f(x)\sqrt{1+\left(f'(x)\right)^2}\,dx \\[4pt]
&= \int_0^4 2\pi \left( -\dfrac{3}{4}x+3\right) \cdot \dfrac{5}{4} \,dx \\[4pt]
&=\dfrac{5pi}{2} \int_0^4 \left( -\dfrac{3}{4}x+3\right) \,dx \\[4pt]
&= \dfrac{5\pi}{2} \left( -\dfrac{3}{8}x^2+3x\right)\bigg|_0^4 \\[4pt] 
&=\dfrac{5\pi}{2}\left[ (-6+12)-(-0+0)\right]  \\[4pt]
&= \boxed{15\pi \text{ units}^2}\end{align*} \]

(b) The surface area when this line is rotated around the \(y\)-axis is given by:

\[\begin{align*} \text{Surface Area} &=\int_a^b 2\pi x\sqrt{1+\left(f'(x)\right)^2}\,dx \\[4pt]
&= \int_0^4 2\pi x \cdot \dfrac{5}{4} \,dx \\[4pt]
&=\dfrac{5\pi}{2} \int_0^4 x\,dx \\[4pt]
&= \dfrac{5\pi}{4} x^2\bigg|_0^4 \\[4pt]  
&= \boxed{20\pi \text{ units}^2}\end{align*} \]

Note that each of these answers can be checked using the surface area of a cone formula from above. The hypotenuse \(h\) is of length \(5\) in each case. When rotating around the \(x\)-axis, the radius \(r\) is \(3\), giving \(A=\pi r h = \pi\cdot 3 \cdot 5=15\pi \). When rotating around the \(y\)-axis, the radius \(r\) is 4, giving \(A=\pi r h = \pi \cdot 4\cdot 5=20\pi\)

Example \(\PageIndex{2}\)

Let \(f(x)=x^3+\dfrac{1}{12x}\) for \(1\le x\le 2\). Find the surface area of the object formed when \(y=f(x)\) is rotated around

a. the \(x\)-axis

b. the \(y\)-axis

Solution

First, we find the derivative of \(f(x)=x^3+\dfrac{1}{12x}\):

\[ f'(x)=3x^2-\dfrac{1}{12x^2} \nonumber \]

Next, we compute \(\sqrt{1+\left(f'(x)\right)^2}\):

\[ \left(f'(x)\right)^2  =\left(3x^2-\dfrac{1}{12x^2}\right)^2  =9x^4-\dfrac{1}{2}+\dfrac{1}{144x^4} \nonumber \]

We see that adding \(1\) to this keeps it a perfect square:

\[ 1+\left(f'(x)\right)^2 =9x^4+\dfrac{1}{2}+\dfrac{1}{144x^4} =\left(3x^2+\dfrac{1}{12x^2}\right)^2 \nonumber  \]

Therefore, inside of each integral we will use

\[ \sqrt{1+\left(f'(x)\right)^2} =3x^2+\dfrac{1}{12x^2} \nonumber \]

(a) The surface area when this curve is rotated around the \(x\)-axis is given by:

\[\begin{align*} \text{Surface Area} &=\int_a^b 2\pi f(x)\sqrt{1+\left(f'(x)\right)^2}\,dx \\[4pt]
&=\int_1^2 2\pi \left(x^3+\dfrac{1}{12x}\right)\left(3x^2+\dfrac{1}{12x^2}\right) dx \\[4pt]
&=2\pi\int_1^2 \left(3x^5+\dfrac{x}{3}+\dfrac{1}{144x^3}\right) dx \\[4pt]
&=2\pi\left[\dfrac{x^6}{2}+\dfrac{x^2}{6}-\dfrac{1}{288x^2}\right]_1^2 \\[4pt]
&=2\pi\left[\left(32+\dfrac{2}{3}-\dfrac{1}{1152}\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}-\dfrac{1}{288}\right)\right] \\[4pt]
&=2\pi\left(32+\dfrac{1}{384}\right) \\[4pt]
&=\boxed{64\pi+\dfrac{\pi}{192}\text{ units}^2} \end{align*}\]

(b) The surface area when this curve is rotated around the \(y\)-axis is given by:

\[ \begin{align*} \text{Surface Area} &=\int_a^b 2\pi x\sqrt{1+\left(f'(x)\right)^2}\,dx \\[4pt]
&=\int_1^2 2\pi x\left(3x^2+\dfrac{1}{12x^2}\right) dx \\[4pt]
&=2\pi\int_1^2 \left(3x^3+\dfrac{1}{12x}\right) dx \\[4pt]
&=2\pi\left[\dfrac{3x^4}{4}+\dfrac{1}{12}\ln (x)\right]_1^2 \\[4pt]
&=2\pi\left[\left(12+\dfrac{1}{12}\ln 2\right)-\left(\dfrac{3}{4}+0\right)\right] \\[4pt]
&=2\pi\left(\dfrac{45}{4}+\dfrac{1}{12}\ln (2)\right) \\[4pt]
&=\boxed{\dfrac{45\pi}{2}+\dfrac{\pi}{6}\ln (2)\text{ units}^2} \end{align*} \]

Exercise \(\PageIndex{1}\)

Find the surface area of the object formed when \(y=\dfrac{1}{8}x^2-\ln(x)\) for \(1\le x\le 4\) is rotated around the \(y\)-axis

Answer

\(\dfrac{33\pi}{2}\)

Not all revolutions are done around an axis. Just like for volumes, if we wish to rotate a curve around a vertical or horizontal line, we can add or subtract the difference in our radius term.

Theorem \(\PageIndex{2}\)

If the curve \(y=f(x)\) for \(a\le x\le b\) is rotated around the line \(y=k\), the surface area of the resulting object is

\[\int_a^b 2\pi \left|f(x)-k\right| \sqrt{1+\left(f'(x)\right)^2}\,dx \nonumber \]

If the curve \(y=f(x)\) for \(a\le x\le b\) is rotated around the line \(x=k\), the surface area of the resulting object is

\[ \int_a^b 2\pi |x-k|\sqrt{1+\left(f'(x)\right)^2}\,dx \nonumber \]

Example \(\PageIndex{3}\)

Find the surface area of the object formed when the curve \(y=\sqrt{1-x^2}\) for \(0\le x\le 1\) is rotated around the line \(x=1\).

Solution

Let \(y = f(x) = \sqrt{1 - x^2}\) has \( f'(x) = \dfrac{-x}{\sqrt{1 - x^2}}\), so we see:

\[ \sqrt{1 + \left(f'(x)\right)^2} = \sqrt{1 + \frac{x^2}{1 - x^2}} = \sqrt{ \dfrac{1}{1 - x^2}}=\dfrac{1}{\sqrt{1-x^2}} \nonumber \] 

The radius when rotating around \(x = 1\) is \(|x - 1|\). Since \(0 \le x \le 1\), we have \(x - 1 \le 0\), so \(|x - 1| = 1 - x\).

The surface area of revolution around \(x=1\) is given by
\[ \begin{align*} \text{Surface Area} &  = \int_a^b 2\pi |x-1| \, \sqrt{1 + \left(f'(x)\right)^2} \, dx \\[4pt]
& = \int_0^1 2\pi\cdot (1-x)\cdot \dfrac{1}{\sqrt{1-x^2}}\,dx \\[4pt] 
&= \lim_{t\to 1^-} 2\pi \int_0^t \dfrac{1-x}{\sqrt{1-x^2}}\,dx \\[4pt]
&= \lim_{t\to 1^-} 2\pi \int_0^t \left(\dfrac{1}{\sqrt{1-x^2}}  - \dfrac{x}{\sqrt{1-x^2}}\right)\,dx \\[4pt] 
&= \lim_{t\to 1^-} 2\pi \left( \sin^{-1}(x)+\sqrt{1-x^2} \right)\bigg|_0^t \\[4pt] 
&=  \lim_{t\to 1^-} 2\pi \left[ \left( \sin^{-1}(t)+\sqrt{1-t^2} \right)-\left( 0+\sqrt{1}\right)\right] \\[4pt]
&= 2\pi \left[ \dfrac{\pi}{2}-1\right] \\[4pt]
&= \boxed{\pi^2 - 2\pi \text{ units}^2} \end{align*} \]

Example \(\PageIndex{4}\)

Find the surface area of the object formed when the curve \(y=\dfrac{1}{18}x^3+4\) for \(0\le x\le 2\sqrt{2}\) is rotated around the line \(y=4\).

Solution

Let \(f(x)=\dfrac{1}{18}x^3+k\) so that \(f'(x)=\dfrac{1}{6}x^2\).

Then

\[ 1+\left(f'(x)\right)^2 = 1+\left(\dfrac{x^2}{6}\right)^2 = 1+\frac{x^4}{36}  \nonumber \]

Notice that since \(x\ge 0\) we have \(f(x)\ge 4\) so \(\left|f(x)-4\right|=f(x)-4=\dfrac{1}{18}x^3\).

The surface area when the curve is rotated around \(y=4\) is given by

\[  \text{Surface Area}  = \int_a^b 2\pi\left| f(x)-4\right| \sqrt{1+\left(f'(x)\right)^2}\,dx =\int_0^{2\sqrt{2}} 2\pi\cdot \dfrac{1}{18}x^3 \sqrt{1+\dfrac{x^4}{36}}\,dx \nonumber  \]

This integral can be solved via substitution, using \(u=1+\dfrac{x^4}{36}\):

\[ \int_0^{2\sqrt{2}} 2\pi\cdot \dfrac{1}{18}x^3 \sqrt{1+\dfrac{x^4}{36}}\,dx =\pi \int_1^{\frac{25}{9}} \sqrt{u}\,du = \dfrac{2\pi}{3}u^{\frac{3}{2}}\bigg|_1^{\frac{25}{9}}=\dfrac{2\pi}{3}\left(\left(\frac{25}{9}\right)^{\frac{3}{2}}-1\right)=\boxed{\dfrac{196\pi}{81}\text{ units}^2} \nonumber  \]

Exercise \(\PageIndex{2}\)

Find the surface area of the object formed when the curve \(y=\dfrac{1}{2}e^x+\dfrac{1}{2}e^{-x}\) for \(0\le x\le \ln(2) \) is rotated around the line \(y=-1\).

Answer

\( \dfrac{15\pi}{8}+2\pi\ln(2)\)

Key Equations

• Surface Area of Revolution for \(y=f(x)\) around \(x\)-axis

\(\displaystyle A=∫^b_a 2\pi f(x)\sqrt{1+\left(f'(x)\right)^2}\,dx\) 

• Surface Area of Revolution for \(y=f(x)\) around \(y\)-axis

\(\displaystyle A=∫^b_a 2\pi x\sqrt{1+\left(f'(x)\right)^2}\,dx\)