How do I evaluate this triple integral using the function integral3

xmin= @(y) y.^2
xmax= @(y) y.^0.5
ymin=0
ymax=1
zmin=0
zmax=@(x,y,z) x+y+36


h = @(y,x,z) 1  %dz dx dy
answer3 = integral3(h, ymin, ymax , xmin , xmax , zmin, zmax);
answer3 = vpa(answer3,8)
%%why doesnt this code work, can someone help me

Answer 1

[Walter Roberson]

xmin= @(y) y.^2

xmin = function_handle with value:
    @(y)y.^2

xmax= @(y) y.^0.5

xmax = function_handle with value:
    @(y)y.^0.5

ymin=0

ymin = 0

ymax=1

ymax = 1

zmin=0

zmin = 0

zmax=@(x,y,z) x+y+36

zmax = function_handle with value:
    @(x,y,z)x+y+36

h = @(y,x,z) ones(size(y))  %dz dx dy

h = function_handle with value:
    @(y,x,z)ones(size(y))

answer3 = integral3(h, ymin, ymax , xmin , xmax , zmin, zmax);
answer3 = vpa(answer3,8)

answer3 = 

12.3

[bob]

what exactly does the ones(size(y)) does? (in context for the integral) I haven't used it

[Walter Roberson]

The integral() family of functions call the given function passing in vectors or arrays of values, expecting the same size of output, using element-wise computations.

So your h(y,x,z) was being called with non-scalar y, x, z, but you were returning the scalar constant 1 no matter what the input size was. You need to return one of those 1's for every input element.

[bob]

oh i see, I understand it now. Thank you