680
Chapter 6 | Applications of Integration
Surface Area = ∫ a b ⎛
⎞ ⎠ dx
⎠ 2
⎝ 2 πf ( x ) 1+ ⎛ 4 ⎛ ⎝ 2 π x 1+ 1 4 x ⎞
⎝ f ′( x ) ⎞
= ∫ = ∫
⎠ dx
1
4 ⎛ ⎝ 2 π x + 1 4 ⎞
1 ⎠ dx . Let u = x +1/4. Then, du = dx . When x =1, u =5/4, and when x =4, u =17/4. This gives us ∫ 1 4 ⎛ ⎝ 2 π x + 1 4 ⎞ ⎠ dx = ∫ 5/4 17/4 2 π udu =2 π ⎡ ⎣ 2 3 u 3/2 ⎤ ⎦ | 5/4 17/4 = π 6 ⎡ ⎣ 17 17−5 5 ⎤ ⎦ ≈30.846.
6.21
Let f ( x ) = 1− x over the interval [0, 1/2]. 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.
Example 6.22 Calculating the Surface Area of a Surface of Revolution 2
Let f ( x ) = y = 3 x 3 .
Consider the portion of the curve where 0≤ y ≤2. Find the surface area of the surface
generated by revolving the graph of f ( x ) around the y -axis.
Solution Notice that we are revolving the curve around the y -axis, and the interval is in terms of y , so we want to rewrite the function as a function of y .Weget x = g ( y ) = (1/3) y 3 . The graph of g ( y ) and the surface of rotation are shown in the following figure.
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