If you've got a curve in 2D, go with Green's theorem. If you've got a curve in 3D, well, Green's theorem breaks down. And how about Green's theorem for curves in higher dimensions? No way, friendo.
But there's a really slick theorem which holds for higher dimensions. Stokes' theorem basically says the same thing as Green's theorem. It's just that we've tagged on a few factors and integral signs, so we can handle higher dimensions.

Basically, the theorem says that

Here, is some curve, and a surface. To apply Stokes' theorem, take care that the unit normal vector is oriented in the right direction. Otherwise your answer will be off by a factor of .

Alright, but why does this hotchpotch of and and all make sense? Have a look at the three cases here. Clearly, the curl on the surface matters for orientation.

But we can actually study whichever surface we like. Why? Well, our closed line integral can be broken down into two separate closed line integrals. On the surface, they'll cancel out. So the boundary is the only thing that matters, really.

Calculate

where is described by

and

Solution: is the part of a sphere with radius centered at the point lying above the -plane.

The condition means that the -component of the normal to should be positive.

The boundary of is where the sphere intersects the -plane. When intersect the -plane, it means that . Thus, is described by

We can now use Stoke's theorem to calculate

But observe now that is also the boundary to the disc described by

Thus, we can use Stoke's theorem again

Since the disc is lying in the -plane, its normal is parallel to the -axis

The dot product means that we only care about the -component of , which is

Thus,

The last integral gives us the area of the disc . Finally, we can conclude that