Surface integral of a vector valued function

Question

The value of the surface integral $$ \iint_S(x\hat{i}+y\hat{j})\cdot \hat{n}~dA $$ evaluated over the surface of a cube having sides of length $a$ is ($\hat{n}$ is unit normal vector) \begin{align*} \iint_S(x\hat{i}+y\hat{j})\cdot \hat{n}~dA &= \iiint_V 2 ~dV \ \text{(divergence theorem)}\\ &=2a^3 \end{align*} but the answer is $0$, I don't know where I am doing the mistake.

Answer 1

The answer in the book is wrong. Without loss of generality, just to simplify calculations, let's choose one of the vertices at the origin, with faces perpendicular to the axes at $0$ and $a$. The faces perpendicular to $z$ axis have the normal either $\hat j$ or $-\hat j$, so they will not contribute to the surface integral. Let's choose the faces perpendicular to the $x$ axis. At $x=0$, the normal is $-\hat i$, so the integrand is $-x=0$, therefore no contribution either. Now at $x=a$, the normal is $\hat n=\hat i$, and the integrand is $x=a$. Integral over the area of the side will yield $a\cdot a^2=a^3$. You can repeat the same for $y$ axis, with contributions $0$ and $a^3$, so the final integral will be the sum of contributions at $x=0$, $x=a$, $y=0$, $y=a$, $z=0$, $z=a$: $$ \iint_S(x\hat{i}+y\hat{j})\cdot \hat{n}~dA =0+a^3+0+a^3+0+0=2a^3$$