Effacer les filtres
Effacer les filtres

How to find surface area ?

4 vues (au cours des 30 derniers jours)
Teerapong Poltue
Teerapong Poltue le 10 Nov 2020
Commenté : Walter Roberson le 10 Nov 2020
how can I find a surface area of this function on certain interval.
f = @(x,y,z) cos(x) + cos(y) + cos(z);

Réponse acceptée

Walter Roberson
Walter Roberson le 10 Nov 2020
So probably something close to
f = @(x,y,z) cos(x) + cos(y) + cos(z);
syms x y z real
syms xl xh yl yh zl zh real
F = f(x,y,z);
SA = int(int(int( sqrt( 1 + diff(F,x).^2 + diff(F,y).^2 + diff(F,z).^2), z, zl, zh), y, yl, yh), x, xl, xh);
disp( char(simplify(SA)) )
int(int((ellipticE(zh, 1/(cos(x)^2 + cos(y)^2 - 3)) - ellipticE(zl, 1/(cos(x)^2 + cos(y)^2 - 3)))*(3 - cos(y)^2 - cos(x)^2)^(1/2), y, yl, yh), x, xl, xh)
  2 commentaires
Teerapong Poltue
Teerapong Poltue le 10 Nov 2020
it isn't work.
I couldn't get from a certain interval as I expected.
Walter Roberson
Walter Roberson le 10 Nov 2020
you would substitute your interval bounds for xl xh yl yh or zl zh
Note that there probably is no closed form so you might need to vpa()

Connectez-vous pour commenter.

Plus de réponses (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by