Why is the interval for theta only pi?

[lumpyduster]

Homework Statement

Verfify Stokes' theorem for the given surface S and boundary ∂S, and vector fields F.

S = [(x,y,z): x²+y²+z²=1, z≥0
∂S = [(x,y): x²+y²=1
F=<x,y,z>

I did this problem and checked the answer - I chose the wrong integral bounds and I am wondering why they are wrong.

Homework Equations

Stokes' theorem:
∫∫(∇×F)dS = ∫F⋅ds

The Attempt at a Solution

I was able to do the left hand of the solution easily. I knew that I should get zero for the cross product, so that's all I had to do.

For the right hand side:

∫F⋅dS = ∫xdx+ydy

x= cosΘ
y=sinΘ
dx=-sinΘ
dy=cosΘ

0≤Θ≤2π

Apparently this is the correct range for theta:

0≤θ≤π

Why is it just π?

 

[vela]

Homework Statement

Verfify Stokes' theorem for the given surface S and boundary ∂S, and vector fields F.

S = [(x,y,z): x²+y²+z²=1, z≥0
∂S = [(x,y): x²+y²=1
F=<x,y,z>

I did this problem and checked the answer - I chose the wrong integral bounds and I am wondering why they are wrong.

Homework Equations

Stokes' theorem:
∫∫(∇×F)dS = ∫F⋅ds

The Attempt at a Solution

I was able to do the left hand of the solution easily. I knew that I should get zero for the cross product, so that's all I had to do.

For the right hand side:

∫F⋅dS = ∫xdx+ydy

x= cosΘ
y=sinΘ
dx=-sinΘ
dy=cosΘ

You should write $dx=-\sin\theta\,d\theta$ and $dy=\cos\theta\,d\theta$.

0≤Θ≤2π

Apparently this is the correct range for theta:

0≤θ≤π

Why is it just π?

0 to $2\pi$ is the correct interval.