Divergence Theorem:
The divergence theorem states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of the vector field over the region enclosed by the surface. Mathematically, it can be written as:

∬(S) F · dA = ∭(V) ∇ · F dV

where,
∬(S) represents the surface integral over the closed surface S,
F is the vector field,
dA is the differential area vector on the surface,
∭(V) represents the volume integral over the region V enclosed by the surface,
∇ · F is the divergence of the vector field F,
and dV is the differential volume element.

Given Vector Field:
The given vector field is F = (ez, sin x, y^2).

Calculating Divergence:
To find the divergence of the vector field, we need to take the dot product of the del operator (∇) with the vector field F.

∇ · F = (∂/∂x, ∂/∂y, ∂/∂z) · (ez, sin x, y^2)
= ∂/∂x (ez) + ∂/∂y (sin x) + ∂/∂z (y^2)
= 0 + cos x + 0
= cos x

Applying the Divergence Theorem:
Now we can apply the divergence theorem to find the flux of the vector field through a closed surface.

The surface integral is given by:
∬(S) F · dA = ∭(V) ∇ · F dV

Since the given function F does not depend on z, the surface integral is independent of z. Therefore, we can choose a surface S that lies in the xy-plane, with z = 0.

Choosing the Surface:
Let's choose a surface S that is a disk in the xy-plane with radius R. The surface S can be parameterized as:
r(u, v) = (R cos u, R sin u, 0)
where 0 ≤ u ≤ 2π and 0 ≤ v ≤ R.

Calculating the Surface Area Element:
To calculate the surface area element dA, we can take the cross product of the partial derivatives of r(u, v) with respect to u and v.

dA = |∂r/∂u × ∂r/∂v| du dv
= |(-R sin u, R cos u, 0) × (-R cos u, -R sin u, 0)| du dv
= |(0, 0, R^2)| du dv
= R^2 du dv

Calculating the Surface Integral:
Now we can calculate the surface integral over the disk S.

∬(S) F · dA = ∫∫(S) F · dA
= ∫∫(S) (ez, sin x, y^2) · (R^2 du dv)
= R^2 ∫∫(S) (0, sin x, y^2) · (du dv)
= R^2 ∫∫(S) sin x du dv

Integration Bounds:
The integration bounds for the surface integral are:
0 ≤ u ≤