I am having problems trying to verify Stokes' theorem (below) as part of a question.
$$\iint_{S} \text{curl} \vec F \cdot d\vec S=\oint_{c} \vec F \cdot d\vec r$$
The vector field in question is $\vec F=(xy,z^3,xz^2)$ and I am looking over a triangle plane that joins $(0,0,1)$ , $(0,4,1)$ and $(4,0,1). $
Computing the surface integral (over the triangle plane) does not seem to yield the same answer as the closed line integral (taking around the edge of the triangle) so I know I am making a mistake somewhere!
My working is below:
Surface Integral:
Function of plane: $$f(z) = z-1$$
Unit normal to plane:
$$\vec n = \frac{\nabla f}{\mod{(\nabla f)}} = (0,0,1)$$
Also,
$$dS = dA(0,0,1) \cdot \vec n = dA = dxdy$$
Curl of vector field:
$$ \text{curl} \vec F = (-3z^2,-z^2,-x)$$
Working out the surface integral:
$$ \int \int _ S \text{curl} \vec F \cdot d\vec S = \int \int _ S (-3z^2,-z^2,-x) \cdot (0,0,1) dxdy$$
$$ = \int \int _ S -x dxdy $$
Limits are $x = [0,4]$,$y=[4-x,4]$
$$ = \int^4_0\int^{4-x}_4 -x dydx $$
$$ = \int^4 _0 x^2 dx=\frac{64}{3}$$
Line Integral
Traversing the edges in an anticlockwise direction, I have split the triangle into three lines. $l_1$ from $(0,0,1)$ to $(0,4,1)$ , $l_2$ from $(0,4,1)$ to $(4,0,1)$ and $l_3$ from $(4,0,1)$ to $(0,0,1)$ .
Also,
$$d\vec r = (dx,dy,dz)$$
$l_1$
$$ \int_{l_1}\vec F \cdot d\vec r = \int_{l_1} (xy,z^3,xz^2) \cdot (dx,dy,dz)$$
$$ = \int_{l_1} xydx+z^3dy+xz^2dz $$
Across $l_1$ $x=0$ $dx=0$ $z=1$ $dz=0$ ($y$ varies).
$$ = \int_{l_1} dy $$
$$ = \int_{0}^4 dy $$
$$ = 4 $$
$l_3$
$$ \int_{l_3}\vec F \cdot d\vec r = \int_{l_3} (xy,z^3,xz^2) \cdot (dx,dy,dz)$$
$$ = \int_{l_3} xydx+z^3dy+xz^2dz $$
Across $l_3$ $y=0$ $dy=0$ $z=1$ $dz=0$ ($x$ varies).
$$ = 0 $$
$l_2$
$$ \int_{l_2}\vec F \cdot d\vec r = \int_{l_1} (xy,z^3,xz^2) \cdot (dx,dy,dz)$$
$$ = \int_{l_2} xydx+z^3dy+xz^2dz $$
Across $l_2$ $x$ and $y$ vary, $y=4-x$ $dy=-dx$ $z=1$ $dz=0$ .
$$ = \int_{l_3} x(4-x)dx – \int_{l_3} dx$$
$$ = \int_{0}^4 4x-x^2-1dx $$
$$ = 32-\frac{64}{3}-4 $$
$$ = \frac{20}{3} $$
And so
$$\oint_{c} \vec F \cdot d\vec r = 4 + \frac{20}{3} = \frac{32}{3}$$
Can anyone see where I am going wrong? Hopefully, I am missing something relatively simple!
Thanks!
Best Answer
The surface associated to your triangle is defined by: $\lambda_1\in[0,1],\ \lambda_2\in[0,1]$ such that $\lambda_1+\lambda_2\le 1$ $$ t(\lambda_1,\lambda_2)=\lambda_1(v_2-v_1)+\lambda_2(v_3-v_1)+v1=(4\lambda_2,4\lambda_1,1) $$ the Curvilinear abscissa are: $$ \gamma_1(\lambda_1,\lambda_2)=\frac{\partial t}{\partial_{\lambda_1}}=(0,4,0) $$ $$ \gamma_2(\lambda_1,\lambda_2)=\frac{\partial t}{\partial_{\lambda_2}}=(4,0,0) $$ Thus surface element $dS$ is $$ dS=\gamma_1\wedge \gamma_2 d\lambda_1d\lambda_2=(0,0,-16) d\lambda_1d\lambda_2 $$ So \begin{align} \int\int curl F.dS&=\int_{\lambda_1=0}^1\int_{\lambda_2=0}^{1-\lambda_1} \underbrace{(-3z^2,-z^2,-x)\circ t(\lambda_1,\lambda_2)}_{(-3,-1,-4\lambda_2)} \cdot \underbrace{dS(\lambda_1,\lambda_2)}_{(0,0,-16) d\lambda_1d\lambda_2} \\ &=\int_{\lambda_1=0}^1\int_{\lambda_2=0}^{1-\lambda_1}64\lambda_2d\lambda_1d\lambda_2 \\ &=64\int_{\lambda_1=0}^1\frac{1}{2}(1-\lambda_1)^2d\lambda_1\\&=\frac{32}{3} \end{align}