Using Stokes' Theorem for Hemispherical Surface Area Calculation

[sci-doo]

Homework Statement

Calculate
\int \int _{S}\nabla \times \overline{F} \cdot \hat{N}dS

where \overline{F} = 3y\hat{i} - 2xz\hat{j} + (x^{2}-y^{2})\hat{k} and S is a
hemispherical surface x² + y² + z² = a², z ≥ 0 and \hat{N} is a normal of the surface outwards. Can you use Stokes' theorem?

Homework Equations

I think I can use Stokes' theorem

\int \int _{S}\nabla \times \overline{F} \cdot \hat{N}dS = \oint_{C} \overline{F} \cdot d\overline{r}

The Attempt at a Solution

\oint_{C} \overline{F} \cdot d\overline{r} = \oint_{C} 3y\hat{i} - 2xz\hat{j} + (x^{2}-y^{2})\hat{k} \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k})
= \oint_{C} (3y dx - 2xz dy + (x^{2}-y^{2})dz)

I don't know how to continue!

I should probably integrate closed line C that is the perimeter curve of the surface (circle radius a), but I have no further idea how to do that. I'm new to this subject and I simply don't understand what I could do/write ..

Help is highly appreciated!

 

[HallsofIvy]

Odd, most people learn how to integrate on curves before getting to Stokes' theorem.

The circle of radius a, in the xy-plane can be written in terms of the parameter \theta as
x= a cos(\theta)
y= a sin(\theta)
z= 0.

dx= -a sin(\theta)d\theta
dy= a cos(\theta)d\theta
dz= 0

 

[sci-doo]

Odd, most people learn how to integrate on curves before getting to Stokes' theorem.

The circle of radius a, in the xy-plane can be written in terms of the parameter \theta as
x= a cos(\theta)
y= a sin(\theta)
z= 0.

dx= -a sin(\theta)d\theta
dy= a cos(\theta)d\theta
dz= 0

Ok, I got a solution -3a^2pi

What I did I made
\overline{r}(t) = a cos t + a sin t where 0 \leq t \leq 2 \pi

and then dot product of F(r) and r'(t)

I integrated that from 0 to 2pi with respect to t.

Can I somehow expect a negative result?