Question

∫(x^2+y^2)^1/2ds(积分区域为L,L为圆周x^2+y^2=ax)的积分值

Answer 1

求∮√(x²+y²)ds（其中l为圆周x²+y²=ax）的积分值
解：l：x²-ax+y²=(x-a/2)²+y²-a²/4=0，故得(x-a/2)²+y²=a²/4，这是一个圆心在(a/2，0)，半径r=a/2的圆；故写成参数形式就是：x=(a/2)(1+cos2t)，y=(a/2)sin2t，t∈[-π/2，π/2].
ds=√[(dx/dt)²+(dydt)²]dt=√[(-asin2t)²+(acos2t)²]dt=adt
故∮√(x²+y²)ds=[-π/2，π/2]a∫√[(a²/4)(1+cos2t)²+(a²/4)sin²2t]dt
=[-π/2，π/2](a²/2)∫√(2+2cos2t)dt=[-π/2，π/2][(√2)a²/2]∫√(1+cos2t)dt
=[-π/2，π/2][(√2)a²/2]∫√(2cos²t)dt=[-π/2，π/2](a²)∫costdt
=(a²)sint︱[-π/2，π/2]=a²[sin(π/2)-sin(-π/2)]=2a².

Answer 2

解：设x=(a/2)cosα+(a/2),
y=sinα
(参数法)
原式=∫﹛[
(a/2)cosα+(a/2)
]
^2
+
(sinα)^2
﹜^
(1/2)
*﹛
[（a/2）sinα]^2+[
(a/2)cosα]^2﹜^(1/2)
dα
α是从0到2π
=∫﹛[
(a^2)/2
+(a
^2)(cosα)/2
]^(1/2)
﹜*(a/2)
dα
=(a^2)/4
∫(2+2cosα)^(1/2)
dα
=(a^2)/4∫abs(
cos(α/2)
)
dα
abs是绝对值的意思
=a^2