Use Stoke's Theorem to evaluate $\int_C -y^3dx + x^3dy - z^3dz$ where $C$ is the intersection of the cylinder $x^2 + y^2 = 1$ and the plane $x + y + z = 1$ oriented counterclockwise.
I got that $F = (-y^3, x^3, -z^3)$. My current calculation for $C$ is $z = x^2-x+y^2-y$ which I got from setting the cylinder and plane equal to each other.
How do I solve the rest of this problem, step by step if possible?
On
You are asked to use Stokes, which means that you have to calculate $\iint_D \operatorname{curl} F\,\cdot d\vec S $, where $D$ is some surface that has $C$ as its boundary. The easiest choice for $D$ is probably the part of the plane that is inside the cylinder.
You need the curl of $F$, which is $$ \operatorname{curl} F=(0,0,3(x^2+y^2)). $$ The easiest way to parametrize the $D$ from above is to take $x=r\cos t$, $y=r\sin t$, $z=1-r\cos t-r\sin t$, with $t\in[0,2\pi)$ and $r\in[0,1]$. The normal vector is then $$ (*,*, r). $$ So \begin{align} \int_C -y^3dx + x^3dy - z^3dz&=\int_0^{2\pi}\int_0^1(0,0,3r^2)\cdot(*,*,r)\,dr\,dt =\frac{3\pi}2. \end{align}
On
$\iint \nabla\times F \cdot\ dS$
$n$ is the normal to the surface which since it is a plane is pretty direct.
$(1,1,1)\ dy\ dx$
$\nabla\times F = (0,0, 3x^2 + 3y^2)$
$3\iint x^2+y^2\ dy\ dx$
At this point I would switch to cylindrical.
$3\int_0^{2\pi}\int_0^1 r^3 \ dr\ d\theta\\ \frac {3\pi}{2}$
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.