Verify Stokes' Theorem for $\mathbf{F}=y\mathbf{i}-x\mathbf{j}$, where the surface $S$ is the paraboloid $z=x^{2}+y^{2}$, with $x^{2}+y^{2}=1$, $z=1$, as its boundary $C$.

Figure 72 $C: x^{2} + y^{2} = 1; z =1$

Solution  Figure 72 shows the paraboloid and the circle $C$ in the plane $z=1$. Stokes' Theorem states the following: $$\oint\limits_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r}=\iint\limits_{\kern-3ptS}{\rm{curl}\, \mathbf{F} \,{\bf\cdot}\, \mathbf{n}}\,dS$$

Parametric equations for $C$ are $x=\cos t$, $y=\sin t$, $z=1,$ $0\leq t\leq 2\pi$. We use the parametric equations to find the line integral $\oint_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r}$ for $\mathbf{F}=y\mathbf{i}-x \mathbf{j}$. $$\begin{eqnarray*} \oint_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r}&=&\oint_{C}(y\,dx-x\,dy)=\int_{0}^{2\pi } \left[ \sin t\left( -\sin t\,dt\right) -\cos t\cos t\,dt\right]\\[4pt] &=&-\int_{0}^{2\pi }[\sin ^{2}t+\cos ^{2}t]\,dt=-\int_{0}^{2\pi }dt=-2\pi \end{eqnarray*}$$

To find the surface integral, $\iint\limits_{\kern-3ptS}{\rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n}}\,dS$, we find $\rm{curl}\, \mathbf{F}$ and $\mathbf{n}$: $$\begin{eqnarray*} \rm{curl}\, \mathbf{F} &\mathbf{=}&\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\[5pt] P & Q & R \end{array} \right\vert =\left\vert \begin{array}{@{}c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\[5pt] y & -x & 0 \end{array} \right\vert \\[4pt] &=&\left[ \dfrac{\partial }{\partial y}0-\dfrac{\partial }{\partial z}(-x) \right] \,\mathbf{i}-\left[ \dfrac{\partial }{\partial x}0-\dfrac{\partial }{ \partial z}y\right] \,\mathbf{j}+\left[ \dfrac{\partial }{\partial x}(-x)- \dfrac{\partial }{\partial y}y\right] \mathbf{k}\\[4pt] &=&0\mathbf{i}-0\mathbf{j}-2\mathbf{k}=-2\mathbf{k} \end{eqnarray*}$$

1048

For $z=f(x,y) =x^{2}+y^{2}$, we have $$\begin{eqnarray*} \mathbf{n\,} &=& \tfrac{-f_{x}(x,y)\mathbf{i}-f_{y}(x,y) \mathbf{j\,}+\,\mathbf{k}}{\sqrt{[f_{x}(x,y)]^{2}+[f_{y}(x,y)]^{2}+1}} = \dfrac{-2x\mathbf{i}-2y\mathbf{j}+\mathbf{k}}{\sqrt{ 4x^{2}+4y^{2}+1}}\\[4pt] \rm{curl}\, \mathbf{F}\,{\bf\cdot}\, \mathbf{n} &=&-2\mathbf{k\,{\bf\cdot}\, }\dfrac{1}{\sqrt{ 4x^{2}+4y^{2}+1}}\left( -2x\mathbf{i}-2y\mathbf{j}+\mathbf{k}\right) =\dfrac{ -2}{\sqrt{4x^{2}+4y^{2}+1}} \end{eqnarray*}$$ So,

where $D$ is the interior of the circle $x^{2}+y^{2}=1$.