Let $\displaystyle \vec{F} = y\vec{i} + z\vec{j} + x\vec{k}$ and C be the boundary of $\displaystyle z = 5 - (x^2 + y^2)$ for z ≥ 3, oriented upward. Find $\displaystyle \int\limits_c\vec{F} · d\vec{r} $.
So to use stokes theorem we need to find the curl of the vector
$\displaystyle \begin{vmatrix}
\vec i & \vec j & \vec k \\
\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\
y & z & x
\end{vmatrix} = (0-1)\vec i-(1-0)\vec j+(0-1)\vec k=-\vec i -\vec j -\vec k$
Now we need the normal vector to the surfe
$\displaystyle F(x,y,z) = x^2+y^2+z-5$
$\displaystyle \nabla F = 2x \vec i + 2y \vec j + \vec k$
$\displaystyle |\nabla F| = \sqrt{4x^2 + 4y^2 + 1} $
$\displaystyle dS=\sqrt{1+(z_x)^2+(z_y)^2}= \sqrt{1+4x^2+4y^2}dA$
$\displaystyle \oint \vec F \cdot d\vec r = \iint_S \nabla \times \vec F\cdot d \vec S= \iint_S \nabla \times \vec F\cdot \frac{\nabla \vec F}{|\nabla F|}\sqrt{1+(z_x)^2+(z_y)^2}dA$
$\displaystyle -\iint_S(2x+2y+1)dA$
We need to integrate over the circle $\displaystyle x^2+y^2=(\sqrt{2})^2$
Just change to polar coordinates to finish
$\displaystyle - \int_0^{2\pi}\int_0^{\sqrt(2)} (2r^2cos\theta + 2r^2\sin\theta + r) dr d\theta=-\int_{0}^{2\pi} \frac{2}{3}r^3\cos(\theta)+\frac{2}{3}r^3\sin(\the ta)+\frac{1}{2}r^2 \bigg|_{0}^{\sqrt{2}}d\theta$
$\displaystyle =-\int_{0}^{2\pi} \frac{4\sqrt{2}}{3}\cos(\theta)+\frac{4\sqrt{2}}{3 }\sin(\theta)+1 d\theta = $
$\displaystyle -\left(\frac{4\sqrt{2}}{3}\sin(\theta)-\frac{4\sqrt{2}}{3}\cos(\theta)+\theta \bigg|_{0}^{2\pi}\right)=-2\pi$
write the function in the form
$\displaystyle F(x,y,z)=0$
$\displaystyle z=9-x^2 \iff x^2+z-9=0$
This gives
$\displaystyle F(x,y,z)=x^2+z-9$
$\displaystyle \nabla F = \frac{\partial F}{\partial x} \vec i+ \frac{\partial F}{\partial y} \vec j+ \frac{\partial F}{\partial z} \vec k =2x\vec i +\vec k$
$\displaystyle d\vec S=\vec n \cdot dS$
Where n is a normal unit vector from the surface
$\displaystyle \vec n =\frac{\nabla F}{|\nabla F|}$
This will always give you a normal unti vector.
Note that IF z can be isolated on your surface then ie
$\displaystyle z=g(x,y)$
then $\displaystyle |\nabla F|=dS=\sqrt{1+(z_x)^2+(z_y)^2}$
and then
$\displaystyle \iint_S F \cdot d\vec S = \iint F \cdot \vec n dA$