运用matlab计算椭圆x^2/4+y^2=1的周长的方法

半椭圆
y=√（1-x^2/4）
椭圆周长=2×∫√（1+y′²）dx
matlab
>> syms x
>> y=(1-x^2/4)^(1/2)
y =
(1 - x^2/4)^(1/2)
>> y1=diff(y,x)
y1 =
-x/(4*(1 - x^2/4)^(1/2))
>> f=(1+y1^2)^(1/2)
f =
(1 - x^2/(16*(x^2/4 - 1)))^(1/2)
>> ss=@(x)(1 - x.^2./(16*(x.^2/4 - 1))).^(1/2)
ss =
@(x)(1-x.^2./(16*(x.^2/4-1))).^(1/2)
>> S=2*quad(ss,-2,2)
S =
9.6885