From Stoke's Theorem: \begin{equation*} \oint_c \textbf{F}\centerdot d\textbf{r}= \int\int_S (\nabla \times\textbf{F})\centerdot d \textbf{S}\end{equation*} Evaluate $\oint _C \textbf{F}\centerdot d\textbf{r}$ by Stoke's Theorem if $\textbf{F}=<1,-2x,3>$ and C: $x^2+y^2=1$.
Choose S to be the hemisphere $x^2+y^2+z^2=1$, where $z \geq 0$.
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The above example is the only one given to us on Stoke's Theorem by our lecturer. I am only able to solve it without using Stoke's Theorem, as I am unsure how to go about solving the double integral of the sphere in order to use Stoke's Theorem.
Can anybody please give me some advice as to how to go about solving this integral using Stoke's? It is VERY important that I get this example, as it is the only one that I will have to reference when practising other examples for my multivariable examination coming up in two week's time.
By Stokes' Theorem
$$\oint_{C}\mathbf{F} \cdot d\mathbf{r} = \int_{S} (\nabla \times \mathbf{F}) \cdot \mathbf{n}dS$$
where the surface $S$ is defined by $z = h(x,y) = \sqrt{1-x^2-y^2}$ with $-1\leq x \leq 1$ and $-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}.$
We have $\mathbf{F}=(1,-2x,3)$ and
$$\nabla \times \mathbf{F} = (0,0,-2),$$
and the outward unit normal vector at $S$ is
$$\mathbf{n}=\frac{(-h_x,-h_y,1)}{\sqrt{1+h_x^2 + h_y^2}}.$$
The differential element of surface area is
$$dS={\sqrt{1+h_x^2 + h_y^2}}dxdy.$$
Hence,
$$\int_{S} (\nabla \times \mathbf{F}) \cdot \mathbf{n}dS = 2\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}}(-2)dxdy= -2\pi$$