Htykuut

I am having problems trying to verify Stokes' theorem (below) as part of a question.

$$iint_{S} text{curl} vec F cdot dvec S=oint_{c} vec F cdot dvec r$$

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:

$$ int int _ S text{curl} vec F cdot dvec S = int int _ S (-3z^2,-z^2,-x) cdot (0,0,1) dxdy$$
$$ = int int _ S -x dxdy $$

Limits are $x = [0,4]$,$y=[4-x,4]$

$$ = int^4_0int^{4-x}_4 -x dydx $$

$$ = int^4 _0 x^2 dx=frac{64}{3}$$

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,

$$dvec r = (dx,dy,dz)$$

$l_1$

$$ int_{l_1}vec F cdot dvec r = int_{l_1} (xy,z^3,xz^2) cdot (dx,dy,dz)$$

$$ = int_{l_1} xydx+z^3dy+xz^2dz $$

Across $l_1$ $x=0$ $dx=0$ $z=1$ $dz=0$ ($y$ varies).

$$ = int_{l_1} dy $$

$$ = int_{0}^4 dy $$

$$ = 4 $$

$l_3$

$$ int_{l_3}vec F cdot dvec r = int_{l_3} (xy,z^3,xz^2) cdot (dx,dy,dz)$$

$$ = int_{l_3} xydx+z^3dy+xz^2dz $$

Across $l_3$ $y=0$ $dy=0$ $z=1$ $dz=0$ ($x$ varies).

$$ = 0 $$

$l_2$

$$ int_{l_2}vec F cdot dvec r = int_{l_1} (xy,z^3,xz^2) cdot (dx,dy,dz)$$

$$ = int_{l_2} xydx+z^3dy+xz^2dz $$

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
$$oint_{c} vec F cdot dvec r = 4 + frac{20}{3} = frac{32}{3}$$

Can anyone see where I am going wrong? Hopefully, I am missing something relatively simple!

Thanks!

The surface associated to your triangle is defined by: $lambda_1in[0,1], lambda_2in[0,1]$ such that $lambda_1+lambda_2le 1$
$$
t(lambda_1,lambda_2)=lambda_1(v_2-v_1)+lambda_2(v_3-v_1)+v1=(4lambda_2,4lambda_1,1)
$$
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
$$
dS=gamma_1wedge gamma_2 dlambda_1dlambda_2=(0,0,-16) dlambda_1dlambda_2
$$
So
begin{align}
intint curl F.dS&=int_{lambda_1=0}^1int_{lambda_2=0}^{1-lambda_1} underbrace{(-3z^2,-z^2,-x)circ t(lambda_1,lambda_2)}_{(-3,-1,-4lambda_2)} cdot underbrace{dS(lambda_1,lambda_2)}_{(0,0,-16) dlambda_1dlambda_2} \
&=int_{lambda_1=0}^1int_{lambda_2=0}^{1-lambda_1}64lambda_2dlambda_1dlambda_2 \ &=64int_{lambda_1=0}^1frac{1}{2}(1-lambda_1)^2dlambda_1\&=frac{32}{3}
end{align}