∫(x^2+y^2)^(1⼀2)ds,其中L为圆x^2+y^2=4x的一周

x－2＝2cost，y＝2sint，t位于【－pi，pi】，dx＝－2sintdt，dy＝2costdt，ds＝根号【(dx)^2+(dy)^2】＝2dt，于是积分变为
积分（从－pi到pi）根号(4(2+2cost))*2dt
=4积分（从－pi到pi）根号(4cos^2(t/2))dt
＝16积分（从0到pi）cos(t/2)dt
=32sin(t/2)|上限pi下限0
＝32

∫√(x²+y²)ds,其中L为圆x²+y²=4x的一周
解：由x²+y²=4x，得x²-4x+y²=(x-2)²+y²=4，故可设x=2+2cost，y=2sint，dx=-2sintdt，dy=2costdt
ds=√(dx²+dy²)=[√(4sin²t+4cos²t)]dt=2dt
故‹L›∫√(x²+y²)ds=[-π，π]∫{√[(2+2cost)²+(2sint)²]}2dt=[-π，π]2∫[√(8+8cost)]dt
=[-π，π]4(√2)∫[√(1+cost)]dt=[-π，π]4(√2)∫√[2cos²(t/2)]dt=[-π，π]8∫cos(t/2)dt
=[-π，π]16∫cos(t/2)d(t/2)=16[sin(t/2)]︱[-π，π]=16[sin(π/2)-sin(-π/2)]=32

x^2+y^2=4x化为标准型为：(x-2)²+y²=4，
则参数方程为：x=2+2cost，y=2sint，t：-π-->π，ds=√[(x')²+(y')²]dt=2dt
则：∫(x²+y²)^(1/2)ds
=∫[-π-->π](4+8cost+4cos²t+4sin²t)^(1/2)*2dt
=∫[-π-->π](8+8cost)^(1/2)*2dt
=4∫[0-->π](8+8cost)^(1/2)dt
=4∫[0-->π](16cos²(t/2))^(1/2)dt
=16∫[0-->π](cos²(t/2))^(1/2)dt
=16∫[0-->π]cos(t/2)dt
=32sin(t/2)
[0-->π]
=32