How do I evaluate this triple integral using the function integral3

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bob 2020-11-10

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评论： bob 2020-11-10

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

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Answer 1

Walter Roberson 2020-11-10

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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

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Walter Roberson 2020-11-10

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 2020-11-10