When the integrand of a surface integral is constant does that mean surface is sphere?

Question

In the surface integral

$$\int \int_S \mathbf{y\cdot n} \:\:\mathrm dS$$

(where $\mathbf{n}$ is an outward pointing normal vector)

1) Is it possible for the elements of $\mathbf{y}$ and $\mathbf{n}$ to vary across the surface but for their dot product to remain constant? I.e. $\mathbf{y\cdot n}=c$ where $c$ is constant? And if yes, then does that means the first equation resolves to

$$\int \int_S \mathbf{y\cdot n} \:\:\mathrm dS=c\int \int_S \:\:\mathrm dS=cA$$

where $A$ is the surface area?

2) If $\mathbf{y\cdot n}=c$, then does that mean the closed surface is a sphere?

3) If not, then: If $\mathbf{y\cdot n}=c$ and the elements of vectors $\mathbf{y}$ and $\mathbf{n}$ do not vary, does that mean the closed surface is a sphere?

Answer 1

1) Yes, you can have ${\bf y}$ and ${\bf n}$ varying across a surface and have the dot product a constant. The trivial case is ${\bf y}=0$. Another example is the electric field of a point-like charge at some distance $r$. The electric field is always pointing along the normal through a sphere of radius $r$. Both ${\bf y}$ and ${\bf n}$ change direction (magnitudes are the same)

2) No. I can define for example the close surface to be a cube, and to define ${\bf y}$ such as ${\bf y}\cdot{\bf n}$ is constant on any face of the cube. Once again, the trivial case is ${\bf y}=0$.

3) Even on a closed sphere, ${\bf n}$ varies in orientation.