✔ 最佳答案
V = ∫ab 2πx f(x) dx
V = ∫0√6 2πx * x2 dx
V = 2π ∫0√6 x3 dx
V = πx4/2 |0√6
V = π/2 * ( 36 – 0 )
V = 18π
http://tw.myblog.yahoo.com/jw!JCVbjQyaBRbXTWOakincl1.Wpxbobg--/article?mid=9181
2011-05-18 11:31:46 補充:
y = x2 ⇒ y' = 2x
V = ∫0√6 2πx * √(1+(y')2) dx
V = ∫0√6 2πx * √(1+(2x)2) dx
V = π ∫0√6 x *(1+4x2)1/2 dx
V = π/4 ∫0√6 4x * (1+4x2)1/2 dx
V = π/4 ∫0√6 (1+4x2)1/2 d(1+4x2)
V = π/6 (1+4x2)3/2 |0√6
V = (π/6) * ( 125 – 1 )
V = 62π/3
因此, 答案為 (D) 62π/3
2011-05-18 11:32:17 補充:
y = x2 ⇒ y' = 2x
V = ∫0√6 2πx * √(1+(y')2) dx
V = ∫0√6 2πx * √(1+(2x)2) dx
V = π ∫0√6 x *(1+4x2)1/2 dx
V = π/4 ∫0√6 4x * (1+4x2)1/2 dx
V = π/4 ∫0√6 (1+4x2)1/2 d(1+4x2)
V = π/6 (1+4x2)3/2 |0√6
V = (π/6) * ( 125 – 1 )
V = 62π/3
因此, 答案為 (D) 62π/3
2011-05-18 17:47:10 補充:
因為要先求曲線之弧長
y = f(x), x ∈ [a,b] 之弧長公式為 ~ ∫ √(1+(f'(x))²) dx (積分範圍為 a ~ b )