Why are these two ways of writing the same integral giving me different results?

function pot1 = potencial_viviani1(R,a)
kappa=8.9875e9;
rho1=3e-5;

f=@(t) R.*sqrt(1+cos(t).^2) ./ norm( [R*cos(t).^2-a(1), R*cos(t).*sin(t)-a(2), R*sin(t)-a(3)] ) ;

pot1=kappa*rho1*integral(f,0,pi);
end

And this alternative way

function pot2 = potencial_viviani2(R,a)
kappa=8.9875e9;
rho1=3e-5;

f= @(t) (R.*sqrt(1+cos(t).^2))./sqrt((R.*cos(t).^2 - a(1)).^2 + (R.*sin(t).*cos(t) -a(2)).^2 + (R.*sin(t) - a(3)).^2);

pot2=kappa*rho1*integral(f,0,pi);
end

I don't know why they're giving me different answers, when I call each function as follows:

potencial_viviani1(2,[2,3,4])
potencial_viviani2(2,[2,3,4])

Answer 1

[G A]

Use the option 'ArrayValued',true

pot1=kappa*rho1*integral(f,0,pi,'ArrayValued',true);

Answer 2

[Jan]

integral() calls the function with a vector as input. Then norm() calcuates the matrix norm in the 1st function, while the 2nd function calculates the vector norm over the 2nd dimension.

x = [1, 2;  3, 4]

x = 2×2

1 2 3 4

norm(x)

ans = 5.4650

sqrt(sum(x .^ 2, 2))

ans = 2×1

2.2361 5.0000

But if you choose vecnorm instead, the difference is still there:

R = 2;
a = [2,3,4];
kappa = 8.9875e9;
rho1  = 3e-5;

% Omit the equal part: R .* sqrt(1 + cos(t) .^ 2) 
f1 = @(t) vecnorm([R * cos(t) .^ 2 - a(1), ...
                   R * sin(t) .* cos(t) - a(2), ...
                   R * sin(t) - a(3)], 2, 2);

f2 = @(t) sqrt((R .* cos(t) .^ 2 - a(1)) .^ 2 + ...
               (R .* sin(t) .* cos(t) - a(2)) .^ 2 + ...
               (R .* sin(t) - a(3)) .^ 2);

f1(1)

ans = 3.4271

f2(1)

ans = 3.4271

Fine. But:

f1([1, 2])

ans = 5.7751

f2([1, 2])

ans = 1×2

3.4271 4.6483

% Better:
f1 = @(t) vecnorm([R * cos(t(:)) .^ 2 - a(1), ...
                   R * sin(t(:)) .* cos(t(:)) - a(2), ...
                   R * sin(t(:)) - a(3)], 2, 2);
f1([1, 2])

ans = 2×1

3.4271 4.6483

Now the values are equal, but the dimensions of intput and output differs, which confuses integral:

Error using integralCalc/finalInputChecks (line 526)
Output of the function must be the same size as the input. If FUN
is an array-valued integrand, set the 'ArrayValued' option to true.

So the output has to be adjusted to the size of the input:

f     = @(t) reshape(R .* sqrt(1 + cos(t(:)) .^ 2) ./ ...
   vecnorm([R * cos(t(:)) .^ 2 - a(1), ...
         R * sin(t(:)) .* cos(t(:)) - a(2), ...
         R * sin(t(:)) - a(3)], 2, 2), size(t));

This is ugly.

Setting the 'ArrayValued' option to true would solve the problem, because then the function is called with a scalar input onle. But you've asked for the reason of the difference. And with the modification, the integration is faster, because it can call f() with an array as input.