# Evaluating the surface integral $\int \nabla \times FdS$ where $F = y\hat{i} + z\hat{j} + x\hat{k}$

Evaluate the surface integral

$$\int \nabla \times FdS$$ where $$F = y\hat{i} + z\hat{j} + x\hat{k}$$

and $$S$$ is the hemisphere $$x^2 + y^2 + z^2 =1$$, $$y\ge 0$$oriented in the direction of positive $$y$$ axis.

Now, I want to solve this with help of divergence theorem so first If I consider the closed surface (ie include the bottom part $$y= 0$$)

Then by Divergence theorem given integral is zero.

So, $$\displaystyle\int_{S1} \nabla \times FdS$$ = - $$\displaystyle\int_{S2} \nabla \times FdS$$

where $$S_2$$ is the surface $$y = 0$$ or $$x^2 + z^2 =1$$,

The normal vector for $$S_2$$ is given a $$-\hat{j}$$

Using this answer to this problem is $$-\pi$$

But my question is, If I use stoke's theorem on the same question then

this is equivalent to solving the line integral

$$\displaystyle\int_{C}{F.dr}$$ where $$C$$ is the curve $$x^2 + z^2 =1$$

and when I solve this by using the parametrization $$x = cost$$, $$z = sin t$$, $$t$$ from $$0$$ to $$2\pi$$ then I get answer as $$\pi$$.

Can anyone tell me why am I getting two different answers to the same problem ? and which one is correct ?

Thank you.

This is an issue of orientation: a bounded surface (the disk $$x^2 + z^2 = 1$$ in your case) and it's boundary have the same orientation if they follow the right-hand rule. When you curl your fingers in the direction of the boundary, your thumb will point in the same direction of the normal vector orienting the surface. So in your case, the parameterization you offer is directed counterclockwise, which by the RHR points your thumb upward (in the direction of $$+ \hat{j}$$). Hence, the orientations do not agree, which is why you are getting answers off by a negative.