计算曲线积分∫xdx/π+(y-x)dy,其中曲线C为摆线x=a(t-sint) y=a(1-cost)(a>0)上从O(0,0)到A(2πa,0)的一段有向弧
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![计算曲线积分∫xdx/π+(y-x)dy,其中曲线C为摆线x=a(t-sint) y=a(1-cost)(a>0)上从O(0,0)到A(2πa,0)的一段有向弧](/uploads/image/z/3913646-14-6.jpg?t=%E8%AE%A1%E7%AE%97%E6%9B%B2%E7%BA%BF%E7%A7%AF%E5%88%86%E2%88%ABxdx%2F%CF%80%2B%28y-x%29dy%2C%E5%85%B6%E4%B8%AD%E6%9B%B2%E7%BA%BFC%E4%B8%BA%E6%91%86%E7%BA%BFx%3Da%28t-sint%29+y%3Da%281-cost%29%28a%3E0%29%E4%B8%8A%E4%BB%8EO%280%2C0%29%E5%88%B0A%282%CF%80a%2C0%29%E7%9A%84%E4%B8%80%E6%AE%B5%E6%9C%89%E5%90%91%E5%BC%A7)
计算曲线积分∫xdx/π+(y-x)dy,其中曲线C为摆线x=a(t-sint) y=a(1-cost)(a>0)上从O(0,0)到A(2πa,0)的一段有向弧
计算曲线积分∫xdx/π+(y-x)dy,其中曲线C为摆线x=a(t-sint) y=a(1-cost)(a>0)上从O(0,0)到A(2πa,0)
的一段有向弧
计算曲线积分∫xdx/π+(y-x)dy,其中曲线C为摆线x=a(t-sint) y=a(1-cost)(a>0)上从O(0,0)到A(2πa,0)的一段有向弧
显然t的取值范围就是0到2π
那么
原积分
=∫ xdx /π +(y-x)dy
=∫(0到2π) { a*(t-sint)/π * a(t-sint)' +[a*(1-cost) -a*(t-sint)] *a(1-cost)' } dt
=∫(0到2π) [a*(t-sint)/π * a(1-cost) +(a-a*cost -at +a*sint) *asint] dt
=a²/π *∫ (t-sint) d(t-sint) +a²∫ (1-cost-t +sint)*sint dt
显然∫ (t-sint) d(t-sint) =0.5(t-sint)²
而∫ (1-cost-t +sint)*sint dt=∫ sint -sintcost-t*sint +sin²t dt
显然
∫sintcost=0.5∫sin2tdt= -0.25cos2t
∫t*sint dt= ∫ -t dcost= -t*cost +∫ cost dt= -t*cost +sint
∫sin²t dt=∫ 0.5-0.5cos2t dt=0.5t -0.25sin2t
所以得到
原积分
=a²/π *0.5(t-sint)² +a² *(-cost+0.25cos2t+t*cost -sint+0.5t -0.25sin2t) 代入上下限2π和0
=a²/π *0.5 *4π² +a² *(2π+π)
=5πa²