Something must be a floating point scalar？

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?

Answer 1

[James Tursa]

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)

[Mathidiot Superfacial]

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]

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 )

Answer 2

[Albert Justin]

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

[Walter Roberson]

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