Chapter 6 | Applications of Integration
679
Theorem 6.6: Surface Area of a Surface of Revolution Let f ( x ) be a nonnegative smooth function over the interval ⎡ ⎣ a , b ⎤ formed by revolving the graph of f ( x ) around the x -axis is given by Surface Area = ∫ a b ⎛ ⎞ ⎠ dx . Similarly, let g ( y ) be a nonnegative smooth function over the interval ⎡ ⎣ c , d ⎤ ⎝ 2 πf ( x ) 1+ ⎛ ⎝ f ′( x ) ⎞ ⎠ 2 revolution formed by revolving the graph of g ( y ) around the y -axis is given by Surface Area = ∫ c d ⎛ ⎝ 2 πg ( y ) 1+ ⎛ ⎝ g ′( y ) ⎞ ⎠ 2 ⎞ ⎠ dy .
⎦ . Then, the surface area of the surface of revolution
(6.9)
⎦ . Then, the surface area of the surface of
Example 6.21 Calculating the Surface Area of a Surface of Revolution 1
Let f ( x ) = 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 the following figure.
Figure 6.46 (a) The graph of f ( x ). (b) The surface of revolution.
⎠ 2 =1/(4 x ). Then,
We have f ( x ) = x . Then, f ′( x ) =1/(2 x ) and ⎛
⎝ f ′( x ) ⎞
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