Getting the area of a surface integral from Matlab

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Niklas Kurz
Niklas Kurz 2021-5-5
评论： Star Strider 2021-5-5
I'd like to approve my solution of where is the unit sphere
Therefore I want to calculate where is the parametrization of the unit sphere:
syms phi the
x = cos(phi).*sin(the);
y = sin(phi).*sin(the);
z = cos(the);
density = x.^2;
para = [x;y;z];
dphi = diff(para,phi);
dthe = diff(para,the);
c = cross(dphi,dthe);
int(int(density*norm(c),phi,0,2*pi),the,0,pi)
Like in my previous quesion I get a cryptic answere, not really any helpfull probably because norm(c) is overcomplicated
Also is there another way to approve the solution. e.g. with trapz?

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Star Strider
Star Strider 2021-5-5
Ran in:
See if simplifying it does what you want —
syms phi the
x = cos(phi).*sin(the);
y = sin(phi).*sin(the);
z = cos(the);
density = x.^2;
para = [x;y;z];
dphi = diff(para,phi);
dthe = diff(para,the);
c = cross(dphi,dthe);
Int2 = int(int(density*norm(c),phi,0,2*pi),the,0,pi)
Int2 = 
Int2 = simplify(Int2, 500)
Int2 = 
.
  2 个评论
Niklas Kurz
Niklas Kurz 2021-5-5
That second entry in simplify was unexpected. I should look at the documentation more extensively...
Star Strider
Star Strider 2021-5-5
No worries!
The full documentation would likely suggest —
Int2 = simplify(Int2, 'Steps',500)
Leaving out the 'Steps' name in the name-value pair is a shortened way of specifying it. Other name-value pairs require the name to be specified as well, also if more than one are specified.

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