Stokes' Theorem: surface integral over the lateral surface of a pyramid

Question

Let $S$ be the lateral surface of the pyramid with points $(0,0,0)$, $(1,3,0)$, $(1,3,5)$ and $(0,3,0)$ as shown: Let $\mathbf{F}(x,y,z)=(y,2x,xyz)$ be a vector field. Evaluate the surface integral $$\iint_S \operatorname{curl}( \mathbf{F} ) \cdot \mathbf{n} \,ds $$ Answer: $3/2$

I parametrized the plane that the triangle is laying and found $\operatorname{curl}(\mathbf{F})$. The problem is the domain of integration of the suface integral. If you have a better way to solve the problem, please be my guest.

Answer 1

Add the downwards oriented base triangle $T$ to your surface $S$to obtain the boundary surface $\partial P$ of the pyramid. Gauss' divergence theorem then gives $$\int_S{\rm curl}({\bf F})\cdot{\bf n}\>{\rm d}\omega+\int_T{\rm curl}({\bf F})\cdot{\bf n}\>{\rm d}\omega=\int_{\partial P}{\rm curl}({\bf F})\cdot{\bf n}\>{\rm d}\omega=\int_P{\rm div}\bigl({\rm curl}({\bf F})\bigr)\>{\rm dvol}=0\ ,$$ since ${\rm div}\bigl({\rm curl}({\bf F})\bigr)\equiv0$. It follows that $$\int_S{\rm curl}({\bf F})\cdot{\bf n}\>{\rm d}\omega=-\int_T{\rm curl}({\bf F})\cdot{\bf n}\>{\rm d}\omega=-\int_T (c_1,c_2,c_3)\cdot(0,0,-1)\>{\rm d}\omega\ .$$ Since $c_3=F_{2.1}-F_{1.2}=1$ we obtain as final result the area of $T$, which is $={\displaystyle{3\over2}}$.

Answer 2

Hint: Let $S_1$ be the triangle $(0,0,0)$, $(1,3,0)$, $(1,3,5)$ [it's analogous for others faces]. Applying Stokes theorem, we get $$\int_{S_1} {\rm curl}({\bf F})\cdot {\bf n} \, {\rm d}s = \oint_{\partial S_1} {\bf F}(x) \, {\rm d}x.$$ Hence, you just have to look at the boundary of $S_1$.

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Let's calculate $\oint_{\partial S_1} {\bf F}(x) \, {\rm d}x$:

We can split the path $\partial S_1$ in three paths: $\gamma_1, \gamma_2, \gamma_3$. For $t\in[0,1]$, we take

$\gamma_1(t) = (t,3t,0)$

$\gamma_2(t) = (1,3,0) + (0,0,5t) = (1,3,5t)$

$\gamma_3(t) = (1,3,5) + (-t,-3t,-5t) = (1-t,3-3t,5-5t)$

and calculate the line integral:

\begin{equation*} \begin{split} \oint_{\partial S_1} {\bf F}(x) \, {\rm d}x & = \sum_{i=1}^3 \int_{\gamma_i} {\bf F}(x)\,{\rm d}x \\ & = \sum_{i=1}^3 \int_0^1 {\bf F}(\gamma_i(t))\cdot \gamma'_i(t)\, {\rm d}t, \end{split} \end{equation*}

where

$$\int_0^1 {\bf F}(\gamma_1(t))\cdot\gamma'_1(t) \, {\rm d}t = \int_0^1 9t \, {\rm d}t = \frac{9}{2},$$

$$\int_0^1 {\bf F}(\gamma_2(t))\cdot\gamma'_2(t) \, {\rm d}t = \int_0^1 75t \, {\rm d}t = \frac{75}{2}$$

and

\begin{equation*} \begin{split} \int_0^1 {\bf F}(\gamma_3(t))\cdot\gamma'_3(t) \, {\rm d}t & = \int_0^1 84t-84 \, {\rm d}t \\ & = \frac{84}{2} - 84 = -42. \end{split} \end{equation*}

Then \begin{equation*} \begin{split} \int_{S_1} {\rm curl}({\bf F})\cdot {\bf n} \, {\rm d}s & = \oint_{\partial S_1} {\bf F}(x)\, {\rm d}x \\ & = \frac{9}{2} + \frac{75}{2} - 42 \\ & = 0 \end{split} \end{equation*}