Calculate the surface integral ∬cosh(√ 2x) cos(y) sin(z)dS over the sphere of radius r = 10 centered at (0, 0, π/2 ).

I want to calculate this integral but it seems that if I use spherical coordinates the integrand becomes overly complicated.

After a little more thought I see that the Laplacian of this function is zero, which implies that the surface integral of the normal/radial derivative of this function over the sphere is zero. But then I get stuck with this approach as well.

So, I think I have figured out this question, I failed to realize that since u(x,y,z)=cosh(√2x)cos(y)sin(z) is harmonic, its value at the center of the sphere is equal to the average of the values that it takes on the surface of the sphere so, with this in mind we can determine that u(0,0,π/2)= 1/(400π)∬cosh(√2x)cos(y)sin(z)dS. So we can conclude that ∬cosh(√2x)cos(y)sin(z)dS = 400π.

• Is the argument of the hyperbolic cosine $\sqrt{2x}$ or $\sqrt{2}\cdot x$? Also, consider using cylindirc coordinates. – Luigi D. Feb 9 '15 at 8:33
• the argument is (√2)⋅x – ACZ Feb 9 '15 at 10:22