The typical transformation for spherical coordinates is

(x,y,z) = (rsinθcosφ, r sinθ­sinφ, rcosθ) and dS = r²sinθ­dθ­dφ

And the full range for a sphere is typically 0 ≤ θ ≤ pi and 0 ≤ φ ≤ 2pi.

By drawing your figure, it's a hemisphere with its bottom on the x-z plane. This allows you to identify that you have the full range for θ. However, by projecting to the x-y plane, you'll find you have a different range for phi, which is -pi/2 ≤ φ ≤ pi/2. You can also say since you're on the surface of the sphere, r = 2. So your surface integral is

I = 2³ * ∫sin²θdθ * ∫sinφdφ

Your zenith integration yields pi/2 and your azimuthal integration yields zero. Hence your integral is zero.

Alternatively, you may use the divergence theorem.

Identify two regions: a hemispherical bowl and the bottom disk. For the hemispherical bowl, you may use the divergence theorem directly. For the bottom of the disk, notice that this corresponds to your surface term y = 0, hence the surface integral is zero. Evidently this problem is equivalent to integrating over the hemispherical bowl, which permits us to use the divergence theorem.

Fortunately, the two answers agree, as they should.

According to the divergence theorem,

∫∫∫∇ · F dV = ∫∫F · n dS where n is the unit normal vector of our surface. For your problem,

n = (sinθcosφ, sinθ­sinφ, cosθ). Since we know f(x,y,z) = y, then we can identify F · n = y = 2sinθ­sinφ → F = 2y, where I've used y to mean the unit normal in the y direction. Hence ∇ · F = 0 and your integral is zero by the divergence theorem.