Something must be a floating point scalar？

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Mathidiot Superfacial
评论： Walter Roberson 2022-4-10
f=@(x,y) sqrt(9-x.^2-y.^2);
xmax=@(y) sqrt(9-y.^2);
volume=integral2(f,0,xmax,0,3)
But it says XMAX must be a floating point scalar？ What's wrong?

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采纳的回答

James Tursa
James Tursa 2016-3-14
编辑：James Tursa 2016-3-14
The error message seems pretty clear. The x limits must by scalar values. The y limits can be functions of x. Just rearrange things so that is the case. Since f is symmetric with respect to x and y, you can just switch arguments.
integral2(f,0,3,0,xmax)
  2 个评论
Mathidiot Superfacial
Mathidiot Superfacial 2016-3-14
what if f is not symmetrical with respect to x and y? Can you still switch like that? for an integral is like SS f(x,y)dxdy, and say there is the code to evaluate like integral2(f,a,b,c,d) then are a and b always the bounds of the inner integral's variable? Or would a and b still be the bounds of x even when I'm trying to evaluate in reverse order like SS f(x,y)dydx ?
Walter Roberson
Walter Roberson 2016-3-14
编辑：Walter Roberson 2016-3-14
For 2D integrals, theory says that it does not matter which order you evaluate the integration. So define the function handle to be integrated so that the first parameter is the one with fixed bounds and the second parameter is the one with variable bounds. Remember it is not required that x be the first parameter.
f = @(y, x) x.^2 + x.*sin(y).^2;
xmax=@(y) sqrt(9-y.^2);
integral2(f, 0, 3, 0, xmax )

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更多回答（1 个）

Albert Justin
Albert Justin 2022-4-10
Enter the function f(x,y)=@(x,y) x.*y
Enter the outer integral lower limit:0
Enter the outer integral upper limit:a
Enter the inner integral lower limit:@(x) x.^2
Enter the inner integral upper limit:@(x) 2-x
i get the same error
  1 个评论
Walter Roberson
Walter Roberson 2022-4-10
Ran in:
a = 5;
f = @(x,y) x.*y
f = function_handle with value:
@(x,y)x.*y
xmin = 0
xmin = 0
xmax = a
xmax = 5
ymin = @(x) x.^2
ymin = function_handle with value:
@(x)x.^2
ymax = @(x) 2-x
ymax = function_handle with value:
@(x)2-x
integral2(f, xmin, xmax, ymin, ymax)
ans = -1.2823e+03

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