Getting the area of a surface integral from Matlab

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?

Answer 1

[Star Strider]

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 = 

.

[Niklas Kurz]

That second entry in simplify was unexpected. I should look at the documentation more extensively...

[Star Strider]

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.