Evaluate $\displaystyle \iint \vec{F}\cdot ds$ where $\vec{F}= 3x\hat{i} + 2y\hat{j} -5z\hat{k}$ and $S$ is the portion of the $y = x^2 + z^2$ that lies behind $y =1$ oriented in the direction of positive y -axis
I am trying to solve this with the help of Gauss theorem.
So, if I consider the disc $S_2 = (x^2 + z^2 = 1)$ along with $S_1 = (y = x^2 + z^2)$ then surface is closed, so Gauss theorem is applicable on $S = S_1 \cup S_2$
Now, $\displaystyle\iint F\cdot ds = \int \operatorname{div}F \,\mathrm{dv}$
But $\nabla\cdot F= 0$ this gives,
$$\iint_{S_2} F\cdot ds + \iint_{S_1} F\cdot ds = 0$$
Now, for $S_2$ outward normal Vector = $\hat{j}$
So, $\displaystyle\iint_{S_1} F\cdot ds = -\iint (2y)\cdot dx dz$
Solving the integral gives $2\pi$
therefore, $\displaystyle\iint_{S_1} F\cdot ds = -2\pi$
However , answer given to me is $2\pi$ I do not understand why I am getting an extra negative sign.
Can someone please tell my mistake ?
When you closed it off with $S_2$, the $S_1$ with its normal pointing in the positive $y$-direction is the inward normal. So there is an extra negative sign.