To easily calculate ${}_E={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$EdAcos$,$ the closed surface must be taken to be a sphere of radius R and $+$q must be placed at its center. Which one is incorrect?

${}_E={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$EdAcos$=E{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$dAcos$$ because the E magnitude is the same at all points of the sphere if $+q$ is placed at the center.

${}_E=E{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$dAcos$=E{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dA$ because E and n become parallel at every point on the sphere. That means $=0$ at all points of the sphere.

${}_E=E{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dA=[k\frac{q}{R^2}]{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dA=[k\frac{q}{R^2}](R^2)$ where $R^2$ is the area of the greatest circle in the sphere.

${}_E=E{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dA=[k\frac{q}{R^2}]{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}dA=[k\frac{q}{R^2}](4R^2),$ where $4R^2$ is the total area of the sphere.