Chapter 6 | Applications of Integration
681
Figure 6.47 (a) The graph of g ( y ). (b) The surface of revolution.
We have g ( y ) = (1/3) y 3 , so g ′( y ) = y 2 and ⎛
⎞ ⎠ 2 = y 4 . Then
⎝ g ′( y )
Surface Area = ∫ c d ⎛
⎞ ⎠ dy
⎞ ⎠ 2
⎝ 2 πg ( y ) 1+ ⎛
⎝ g ′( y )
2 ⎛
⎞ ⎠ dy
⎛ ⎝ 1 3
y 3 ⎞
= ∫
⎠ 1+ y 4
⎝ 2 π
0
2 ⎛ ⎝ y 3 1+ y 4 ⎞
3 ∫
= 2 π
0 ⎠ dy . Let u = y 4 +1. Then du =4 y 3 dy . When y =0, u =1, and when y =2, u =17. Then 2 π 3 ∫ 0 2 ⎛ ⎝ y 3 1+ y 4 ⎞ ⎠ dy = 2 π 3 ∫ 1 17 1 4 udu
⎦ | 1
⎡ ⎣ 2 3
u 3/2 ⎤
17
⎡ ⎣ (17) 3/2 −1 ⎤
= π 6
= π 9
⎦ ≈24.118.
6.22
Let g ( y ) = 9− y 2 over the interval y ∈ [0, 2]. Find the surface area of the surface generated by revolving the graph of g ( y ) around the y -axis.
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