How to find surface area ?

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Teerapong Poltue
Teerapong Poltue il 10 Nov 2020
Commentato: Walter Roberson il 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);

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

Accedi per commentare.

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