MQ18 --- Surface Integral

2012-06-01 6:03 am
Difficulty: 60%Evaluate the area of thesurface generated by rotating the curve y = sinx between x = 0 and x = π around the x-axis.

回答 (2)

2012-06-01 7:47 am
✔ 最佳答案
我昨天能貼圖

今天卻不能貼圖

怪怪了

請去下列網址觀看

http://img824.imageshack.us/img824/9153/66555691.gif

2012-06-01 17:14:44 補充:
計算過程大致跟樓下老兄差不多

不過他的答案是錯的

我用手打需要一點時間

等等

2012-06-01 17:25:17 補充:
S = ∫( 0→π )2π。sinx。( 1 + cos^2x )^( 1/2 )dx

令cosx = u → du = - sinxdx

= -2π∫( 1 → -1 )( 1 + u^2 )^( 1/2 )du

= 4π∫( 0 → 1 )( 1 + u^2 )^( 1/2 )du

令u = tant → du = sec^2tdt

= 4π∫( 0 → π/4 )sec^3tdt

= 2π[ tant。sect + ln( sect + tant ) ] | ( 0→π/4 )

= 2π[ 2^(1/2) + ln( 2^(1/2) + 1 ) ]
2012-06-01 7:54 am

圖片參考:http://i1191.photobucket.com/albums/z467/robert1973/Jun12/Crazyint1.jpg
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