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?