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Surface Area Help

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By — McGraw-Hill Professional
Updated on Sep 1, 2011

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.

Applications of the Integral 8.5 Arc Length and Surface Area

Fig. 8.31

We partition the interval [ a, b ]:

a = x 0x 1x 2 ≤ ... ≤ x k −1x k = b.

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

Applications of the Integral 8.5 Arc Length and Surface Area

Fig. 8.32

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 j . This length was determined above to be about

(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 .

Applications of the Integral 8.5 Arc Length and Surface Area

Fig. 8.33

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

Applications of the Integral 8.5 Arc Length and Surface Area

But this sum is also a Riemann sum for the integral

Applications of the Integral 8.5 Arc Length and Surface Area

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

Applications of the Integral 8.5 Arc Length and Surface Area

represents the area of the surface of revolution.

Examples

Example 1

Let f ( x ) = 2 x3 . 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

Applications of the Integral 8.5 Arc Length and Surface Area

This integral is easily calculated using the u -substitution u = 36 x4 , du = 144 x3 dx . With this substitution the limits of integration become 36 and 576; the area is thus equal to

Applications of the Integral 8.5 Arc Length and Surface Area

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

Applications of the Integral 8.5 Arc Length and Surface Area

Applications of the Integral 8.5 Arc Length and Surface Area

Fig. 8.34

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

Applications of the Integral 8.5 Arc Length and Surface Area

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

Applications of the Integral 8.5 Arc Length and Surface Area

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

Applications of the Integral 8.5 Arc Length and Surface Area

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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