Solve double integral using 'integral2'

Suppose I have this surface integral:

I use integral2 to solve the double integral but the result has complex number in it

My codes are:

syms x y z 
format rat
x=sqrt(1-y.^2-z.^2)
xy=diff(x,y)
xz=diff(x,z)
dS = sqrt(100 + xy.^2 + xz.^2)
fun1 = subs((x+y+z).*dS)
f = matlabFunction(fun1)
Myz = integral2(f,0,10,0,@(y)sqrt(100-y.^2))

And the answer

   f = 

      @(y,z)(y+z+sqrt(-y.^2-z.^2+1.0)).*sqrt(-y.^2./(y.^2+z.^2-1.0)-z.^2./(y.^2+z.^2-1.0)+1.0e2)

Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test. 
> In integral2Calc>integral2t (line 129)
  In integral2Calc (line 9)
  In integral2 (line 106)
  In Untitled3 (line 18) 

Myz =

    139388/21     +97492/19i

What's the problem guys ? Thank you

Answer 1

[Torsten]

y = sqrt(100-x^2-z^2) or y = -sqrt(100-x^2-z^2)
->

I = integral_{x=0}^{x=10} integral_{z=0}^{z=sqrt(1-x^2)} (x+sqrt(100-x^2-z^2)+z)*sqrt(1+x^2/(100-x^2-z^2)+z^2/(100-x^2-z^2)) dz dx +

integral_{x=0}^{x=10} integral_{z=0}^{z=sqrt(1-x^2)} (x-sqrt(100-x^2-z^2)+z)*sqrt(1+x^2/(100-x^2-z^2)+z^2/(100-x^2-z^2)) dz dx =

integral_{x=0}^{x=10} integral_{z=0}^{z=sqrt(1-x^2)} 2*(x+z)*10/sqrt(100-x^2-z^2) dz dx

In MATLAB:

I = integral2(@(x,z)2*(x+z)*10./sqrt(100-x.^2-z.^2),0,10,0,@(x)sqrt(100-x.^2))

Best wishes

Torsten.