Stoke's Theorem problem (direct calculation and using Stoke's Theorem)

Question

Here are the question and my answer for the i) part. I have no idea how to do ii), but I know it has the same answer as the one in i)

Let $\Gamma$ be the triangle consisting of the line segments $(3,0,0)$ to $(0,4,0)$, from $(0,4,0)$ to $(0,0,5)$ and from $(0,0,5)$ to $(3,0,0)$ Evaluate the line integral: $$\oint_\Gamma ydx+zdy+xdz$$ by (i) direct calculation, (ii) using Stokes' Theorem

(i) Directly: $$C_1=(3,0,0) \rightarrow (0,4,0)$$ $$C_2=(0,4,0) \rightarrow (0,0,5)$$ $$C_3=(0,0,5) \rightarrow (3,0,0)$$

Then,

$$\oint_\Gamma ydx+zdy+xdz = \oint_\Gamma (y,z,x)dr$$

\begin{align}C1: r(t)&=(3,0,0)+t((0,4,0)-(3,0,0)) =(3-3t,4t,0)\newline r'(t)&=(-3,4,0)dt \end{align}

$$\Rightarrow \int_{C1} (4t,0,3-3t)\cdot(-3,4,0)dt = -6$$

\begin{align}C2: r(t)&=(0,4,0)+t((0,0,5)-(0,4,0)) =(0,4-4t,5t)\newline r'(t)&=(0,-4,5)dt \end{align}

$$\Rightarrow \int_{C2} (4-4t,5t,0)\cdot(0,-4,5)dt = -10$$

\begin{align}C3: r(t)&=(0,0,5)+t((3,0,0)-(0,0,5)) =(3t,0,5-5t)\newline r'(t)&=(3,0,-5)dt \end{align}

$$\Rightarrow \int_{C3} (0,5-5t,3t)\cdot(3,0,5)dt = -\frac{15}{2}$$

$$\therefore \oint_\Gamma ydx+zdy+xdz=\int_{C_1}(y,z,x)dr+\int_{C_2}(y,z,x)dr+\int_{C_3}(y,z,x)dr=-23.5$$

Answer 1

Let $T$ be the triangle surface and ${\bf F}(x,y,z):=(y,z,x)$ the given vector field. Then Stokes' theorem says that $$\int_{\partial T}{\bf F}\cdot d{\bf r}=\int_T{\rm curl}({\bf F})\cdot{\bf n}\>{\rm d}\omega\ ,\tag{1}$$ whereby the orientations of $T$ and $\partial T$ are coupled in the usual way. In order to compute the flux integral on the right hand side of $(1)$ one usually has to provide a parametric representation $(u,v)\mapsto{\bf r}(u,v)$ of the surface in question. Then $$\int_T{\rm curl}({\bf F})\cdot{\bf n}\>{\rm d}\omega=\int_{\hat T}{\rm curl}({\bf F})\bigl({\bf r}(u,v)\bigr)\cdot({\bf r}_u\times{\bf r}_v)\>{\rm d}(u,v)\ . $$ But here is a simpler way since (i) ${\rm curl}({\bf F})=(-1,-1,-1)$ is constant and (ii) $T$ is a plane surface, namely half the parallelogram spanned by the vectors ${\bf a}=(-3,4,0)$ and ${\bf b}=(-3,0,5)$. The flux $\Phi$ in question then is just $$\Phi={1\over2}{\rm curl}({\bf F})\cdot({\bf a}\times{\bf b})={1\over2}(-1,-1,-1)\cdot(20,15,12)=-23.5\ .$$

Answer 2

I know this isn't an answer, but I can't add a comment.

Here's a page that explains the theorem and its use http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx