$\iint \nabla \times F\ dA = \oint f\cdot dr$
$\nabla \times F = (\frac{\partial}{\partial y} xy - \frac{\partial}{\partial z} x^2,\frac{\partial}{\partial z} 2y- \frac{\partial}{\partial x} xy,\frac{\partial}{\partial x} x^2 - \frac{\partial}{\partial y} 2y) = (x, -y, 2x-2)$
..... you can skip much of what follows, I am complicating things more than I need to ....
the normal to the surface is $(\frac {1}{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3})$
$\iint \frac {3x - y - 2}{\sqrt 3} \ dS$
Parameterize that surface?
$u = \frac {x-y}{\sqrt 2}\\
v = \frac {x+y - 2z}{\sqrt 6} \\
w = \frac {x+y+z}{\sqrt 3}$
This coordinate transformation is orthonormal.
jacobian $\ dx\ dy\ dz = du\ dv\ dw$
In these coordinates $w = 0$
The semi-major axis of this elipse is at the point $(0,\sqrt {3},0)$
and the semi-minor axis is at $(1,0,0)$
or
$x = \frac{1}{\sqrt 2} u + \frac {1}{\sqrt 6} v + \frac{1}{\sqrt 3} w\\
y = -\frac{1}{\sqrt 2} u + \frac {1}{\sqrt 6} v + \frac{1}{\sqrt 3} w\\
z = -\frac {2}{\sqrt 6} v + \frac{1}{\sqrt 3} w$
$x^2 + y^2 = 1$ with $x+y+z = 0$ becomes $u^2 + \frac {v^2}3 = 1$ with $w= 0$
Putting it together we have
$\iint \frac {a_1 u + a_2 v - 2}{\sqrt 3} \ du \ dv$
We don't need to work out what the linear coefficients of u,v are because whatever they are, over this ellipse:
$\iint a_1 u + a_2 v\ du\ dv = 0$
$-\frac 2{\sqrt 3} A$
$A$ is the area of the ellipse. $A = \pi\sqrt 2$
The integral evaluates to $-2\pi$
.... Resume here, I don't have the heart do delete it .......
instead of the coordinate transformation, we can say:
$x = x\\y = z\\ z= -x-y\\dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y}, 1)\ dx\ dy$
$\iint (x, -y, 2x-2)\cdot (1,1,1)\ dx\ dy\\
\iint x -y+ 2x-2\ dx\ dy$
$\iint x -y+ 2x \ dx\ dy$ are odd functions evaluated over a region that is symmetric.
$\iint -2 \ dx\ dy = -2\pi$
How did this compare to calculating it directly?
$(x,y,z) = (\cos t, \sin t, 1 - \cos t - \sin t)
dr = (-\sin t, \cos t, 1 + \sin t - \cos t)$
$F\cdot dr = (2\sin t, \cos^2 t, \cos t\sin t)\cdot(-\sin t, \cos t, 1 + \sin t - \cos t)\\
-2\sin^2t + \cos^3 t + \cos t\sin t + \cos t\sin^2 t - \cos^2 t\sin t$
That is going to be a snap to integrate. Most of those terms evaluate to 0
$\int_0^{2\pi} -2\sin^2t + \cos^3 t + \cos t\sin t + \cos t\sin^2 t - \cos^2 t\sin t \ dt\\
\int_0^{2\pi} -2\sin^2t\ dt\\
-2\pi$
Parameterize the surface $S:z=1-x-y$ with the vector function $\vec{r}(x,y)=\left\langle x,y,1-x-y\right\rangle$, where $(x,y)$ belongs to the unit disk $D:x^2+y^2\leq 1$.
Compute the normal vector to $S$ as $\vec{r}_{x}\times\vec{r}_{y}=\langle-z_x,-z_y,1\rangle$.
Calculate $\textrm{curl }\vec{F}$ and plug in $\vec{r}(x,y)$ to get $\textrm{(curl }\vec{F})(\vec{r}(x,y))$.
Take the dot product $\textrm{(curl }\vec{F})(\vec{r}(x,y))\cdot (\vec{r}_{x}\times\vec{r}_{y})$.
Evaluate the surface integral $\iint_S \textrm{curl }\vec{F} \cdot d\vec{S}=\iint_D \textrm{(curl }\vec{F})(\vec{r}(x,y))\cdot (\vec{r}_{x}\times\vec{r}_{y})\,dA$ by converting to polar coordinates.
Best Answer
Let $\mathbf{F}=\left(-y^3,x^3,z^3\right).$ So we have \begin{align*} \oint_{C} -y^3dx+x^3dy+z^3dz&=\oint_C\mathbf{F}\cdot d\mathbf{r} \\ &=\iint_S(\nabla\times\mathbf{F})\cdot dS \\ &=\iint_S(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\,dA \\ &=\sqrt{3}\iint_S\left(x^2+y^2\right)\,dA. \end{align*} My hunch is that we can consider the projection into the $xy$ plane of the surface $S$ for this problem, because the integrand $\nabla\times\mathbf{F}$ only has a $z$ component. If that is so, we will want to switch to polar coordinates: \begin{align*} \oint&=\sqrt{3}\int_0^{2\pi}\int_0^1\left(x^2+y^2\right)\,r\,dr\,d\theta \\ &=2\sqrt{3}\,\pi\int_0^1 r^3\,dr \\ &=\frac{\sqrt{3}\,\pi}{2}. \end{align*} As we can see from this wiki, to adjust for the projection, we need $$A_{\text{proj}}=\cos(\beta) A, $$ since the angle is constant and pulls out of the integral. We have computed the projected area, so we must compensate by dividing by $\cos(\beta),$ we can calculate via the dot product formula: $$\frac{1}{\sqrt{3}}(1,1,1)\cdot(0,0,1)=\cos(\beta). $$ This means the final result is $$\frac{\sqrt{3}\,\pi}{2}\div\frac{1}{\sqrt{3}}=\frac{3\pi}{2}. $$
Now the question is, was the projection justified? Can we verify by, say, computing the original line integral? As suggested by DiegoMath in his answer, we can parametrize the curve $C$ as \begin{align*} x&=\cos(t) \\ y&=\sin(t) \\ z&=1-\cos(t)-\sin(t),\\ 0&\le t\le 2\pi. \end{align*} Then we have \begin{align*} \mathbf{r}(t)&=(\cos(t), \sin(t), 1-\cos(t)-\sin(t))\\ \dot{\mathbf{r}}(t)&=(-\sin(t), \cos(t), \sin(t)-\cos(t)) \\ \mathbf{F}(t)&=(-\sin^3(t),\cos^3(t),(1-\cos(t)-\sin(t))^3) \\ \mathbf{F}\cdot\dot{\mathbf{r}}&=\sin^4(t)+\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3 \\ \oint_C\mathbf{F}\cdot d\mathbf{r}&=\oint_C\mathbf{F}\cdot \dot{\mathbf{r}}(t)\,dt \\ &=\int_0^{2\pi}\left[\sin^4(t)+\cos^4(t)+(\sin(t)-\cos(t))(1-\cos(t)-\sin(t))^3\right]dt \\ &=\frac{3\pi}{2}, \end{align*} which is the answer we had above.