How do I evaluate this triple integral using the function integral3
5 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
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
0 commentaires
Réponse acceptée
Walter Roberson
le 10 Nov 2020
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) ones(size(y)) %dz dx dy
answer3 = integral3(h, ymin, ymax , xmin , xmax , zmin, zmax);
answer3 = vpa(answer3,8)
3 commentaires
Walter Roberson
le 10 Nov 2020
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.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Numerical Integration and Differential Equations dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!