How to find surface area ?

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Teerapong Poltue
Teerapong Poltue el 10 de Nov. de 2020
Comentada: Walter Roberson el 10 de Nov. de 2020
how can I find a surface area of this function on certain interval.
f = @(x,y,z) cos(x) + cos(y) + cos(z);

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Walter Roberson
Walter Roberson el 10 de Nov. de 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 comentarios
Teerapong Poltue
Teerapong Poltue el 10 de Nov. de 2020
it isn't work.
I couldn't get from a certain interval as I expected.
Walter Roberson
Walter Roberson el 10 de Nov. de 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()

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