Boundary of a surface z=\sin(x)\cos(y)

Question

I am doing a question on Stoke's theorem, and one of the requirements is to find the boundary of the surface $$z=sin(x)cos(y), 0{\leq}x{\leq}\pi, 0{\leq}y{\leq}\frac{\pi}{2}$$ So far, I think that the boundary is where z=0, and such the lines $x=0,x=\pi, y=\frac{\pi}{2}$, but this doesn't seem to make a closed boundary to integrate over! I've realised I don't have any method to find the boundary except attempting to visualise it, and some educated guessing. Any help would be appreciated, thanks.

Answer 1

That function is defined over a rectangular area, and it is continuous, so there is no boundary of the surface on the inside. The boundary of the surface will be over the bounding rectangle. So for each of the 4 sides of that bounding rectangle, find out what $z$ is.