求∫∫y^2dσ,其中D是由摆线x=a(t-sint),y=a(1-cost)(0≤t≤2π)的一拱与x轴所围成

先积y,
∫∫y²dσ
=∫[0---->2πa] dx∫[0--->y(x)] y²dy
=(1/3)∫[0---->2πa] y³(x) dx
换元：令x=a(t-sint),则y(x)=a(1-cost),dx=a(1-cost)dt,t：0---->2π
=(1/3)∫[0---->2π] a⁴(1-cost)⁴ dt
=(a⁴/3)∫[0---->2π] (1-cost)⁴ dt
=(a⁴/3)∫[0---->2π] (1-4cost+6cos²t-4cos³t+cos⁴t) dt
平方和四次方降幂
=(a⁴/3)∫[0---->2π] (1-4cost+3(1+cos2t)-4cos³t+(1/4)(1+cos2t)²) dt
=(a⁴/3)∫[0---->2π] (1-4cost+3(1+cos2t)-4cos³t+(1/4)(1+2cos2t+cos²2t)) dt
=(a⁴/3)∫[0---->2π] ((17/4)-4cost+(7/2)cos2t-4cos³t+(1/4)cos²2t) dt
再降幂
=(a⁴/3)∫[0---->2π] ((17/4)-4cost+(7/2)cos2t-4cos³t+(1/8)(1+cos4t)) dt
=(a⁴/3)∫[0---->2π] ((35/8)-4cost+(7/2)cos2t+(1/8)cos4t) dt-(4a⁴/3)∫[0---->2π] cos³t dt
=(a⁴/3)[(35/8)t-4sint+(7/4)sin2t+(1/32)sin4t]-(4a⁴/3)∫[0---->2π] cos²t d(sint)
=(a⁴/3)[(35/8)t-4sint+(7/4)sin2t+(1/32)sin4t]-(4a⁴/3)∫[0---->2π] (1-sin²t) d(sint)
=(a⁴/3)[(35/8)t-4sint+(7/4)sin2t+(1/32)sin4t]-(4a⁴/3)(sint-(1/3)sin³t |[0---->2π]
=(a⁴/3)*(35/8)*(2π)
=35πa⁴/12