有大佬知道这个matlab代码哪里错了么？

Question

f=@(x,y)h-3.086*(x^2)-20.4082*(y^2);
q=integral2(f,-1/(30.864)^(1/2),1/(30.864)^(1/2),-1/(204.082)^(1/2),1/(204.082)^(1/2));

Answer 1

1. 需要给h赋值；

2. f函数中针对x和y的^操作需要改成 .^

h = 1;
f=@(x,y) h-3.086*(x.^2)-20.4082*(y.^2);
q=integral2(f,-1/(30.864)^(1/2),1/(30.864)^(1/2),-1/(204.082)^(1/2),1/(204.082)^(1/2))

结果：

q =

    0.0470

 

已知q求h的办法，下面的函数文件存成funq.m文件

function q = funq(h)
f=@(x,y) h-3.086*(x.^2)-20.4082*(y.^2);
q=integral2(f,-1/(30.864)^(1/2),1/(30.864)^(1/2),-1/(204.082)^(1/2),1/(204.082)^(1/2));

求解m脚本文件

q = 0.5;
f = @(h) funq(h)-q;
h0 = 1;
h = fsolve( f,h0  )

结果：

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

<stopping criteria details>

h =

    9.9873

验算：

>> f(h)

ans =

   2.0365e-09

>> funq(h)

ans =

    0.5000

Answer 2

f=@(x,y)h-3.086*(x^2)-20.4082*(y^2);
q=integral2(f,-1/(30.864)^(1/2),1/(30.864)^(1/2),-1/(204.082)^(1/2),1/(204.082)^(1/2));

Answer 3

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