**Introduction to Surface Area**

Let *f* ( *x* ) be a non-negative function on the interval [ *a, b* ]. Imagine rotating the graph of *f* about the *x* -axis. This procedure will generate a surface of revolution, as shown in Fig. 8.31. We will develop a procedure for determining the area of such a surface.

We partition the interval [ *a, b* ]:

*a = x* _{0} ≤ *x* _{1} ≤ *x* _{2} ≤ ... ≤ *x* _{k −1} ≤ *x _{k}* =

*b*.

Corresponding to each pair of elements *x* _{j −1} , *x _{j}* in the partition is a portion of curve,

as shown in Fig. 8.32. When that piece of curve is rotated about the *x* -axis, we obtain a cylindrical surface. Now the area of a true right circular cylinder is 2 *π · r · h* . We do not have a true cylinder, so we proceed as follows. We may approximate the radius by *f* ( *x _{j}* ). And the height of the cylinder can be approximated by the length of the curve spanning the pair

*x*

_{j −1},

*x*. This length was determined above to be about

_{j}(1 + [ *f′* ( *x _{j}* )]

^{2})

^{1/2}Δ

*x*.

_{j}Thus the area contribution of this cylindrical increment of our surface is about

2 π* · f* ( *x _{j}* )(1 + [

*f′*(

*x*)]

_{j}^{2})

^{1/2}Δ

*x*.

_{j}See Fig. 8.33. If we sum up the area contribution from each subinterval of the partition we obtain that the area of our surface of revolution is about

But this sum is also a Riemann sum for the integral

As the mesh of the partition gets finer, the sum (*) more closely approximates what we think of as the area of the surface, but it also converges to the integral.

We conclude that the integral

represents the area of the surface of revolution.

**Examples**

**Example 1**

Let *f* ( *x* ) = 2 *x*^{3} . For 1 ≤ *x* ≤ 2 we rotate the graph of *f* about the *x* -axis. Calculate the resulting surface area.

**Solution 1**

According to our definition, the area is

This integral is easily calculated using the *u* -substitution *u* = 36 *x*^{4} , *du* = 144 *x*^{3} *dx* . With this substitution the limits of integration become 36 and 576; the area is thus equal to

**Example 2**

Find the surface area of a right circular cone with base of radius 4 and height 8.

**Solution 2**

It is convenient to think of such a cone as the surface obtained by rotating the graph of *f* ( *x* ) = *x* /2, 0 ≤ *x* ≤ 8, about the *x* -axis ( Fig. 8.34 ). According to our definition, the surface area of the cone is

**You Try It** : The standard formula for the surface area of a cone is

Derive this formula by the method of Example 8.24.

We may also consider the area of a surface obtained by rotating the graph of a function about the *y* -axis. We do so by using *y* as the independent variable. Here is an example:

**Example 3**

Set up, but do not evaluate, the integral for finding the area of the surface obtained when the graph of *f* ( *x* ) = *x* ^{6} , 1 ≤ *x* ≤ 4, is rotated about the *y* -axis.

**Solution 3**

We think of the curve as the graph of *φ* ( *y* ) = *y* ^{1/6} , 1 ≤ *y* ≤ 4096. Then the formula for surface area is

Calculating *φ′* ( *y* ) and substituting, we find that the desired surface area is the value of the integral

**You Try It** : Write the integral that represents the surface area of a hemisphere of radius one and evaluate it.

Find practice problems and solutions for these concepts at: Applications of the Integral Practice Test.

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