Calculate the work done by the field $F$ along the curve $L$ with anticlockwise orientation where $F(x,y,z)=(\sin (x^2)-y,2x+y+z,\arctan (z)-x)$ and $L=\{(x,y,z)\mid 2x+4y+3z=0,x^2+y^2+z^2=1\}$.
My thoughts:$$\nabla \times F=(-1,1,3),$$which motivates the use of Stokes' Theorem,$$\int \limits _LF\cdot dr=\iint \limits _S(\nabla \times F)\cdot n\,dS,$$where $S$ is a surface such that $\partial S=L$.
I decided to work with$$S=\{(x,y,z)\mid 2x+4y+3z=0,x^2+y^2+z^2\leq 1\},$$with an upward facing normal. My question is:
How can I parameterize $S$?
My attempt: Taking $z=\frac{-2x-4y}{3}$ implies $x^2+y^2+\left (\frac{-2x-4y}{3}\right )^2\leq 1$, which is equivalent to $13x^2+16xy+25y^2\leq 9$. This implies that a possible parameterization is$$r(x,y)=\left (x,y,\frac{-2x-4y}{3} \right),\,\,13x^2+16xy+25y^2\leq 9.$$My main issue is with the parameterization of $13x^2+16xy+25y^2\leq 9$ in the $xy$ plane. This is a rotated ellipse, and I am not sure what might be the best way to approach this.
Thanks!
So first off, a correction. Stokes' Theorem has a dot product, not a cross product: $$ \oint \mathbf{F}\cdot d\mathbf{r} = \iint(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\,dS. $$
Anyways, you're overcomplicating things. Let's think about the geometry of the problem. Since S is contained in a plane, $\hat{n}$ is constant, and in particular is equal to $(2,4,3)/\sqrt{2^2 + 4^2 + 3^2} = (2,4,3)/\sqrt{29}$. So $(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}} = (-1,1,3)\cdot(2,4,3)/\sqrt{29} = 11/\sqrt{29}$. The surface integral of a constant is just that constant times the area of the surface. What's the area of the surface? Well, the intersection of any plane through the origin and the unit sphere is a unit circle. So the area of $S$ is $\pi$. Thus we have $$ \oint \mathbf{F}\cdot d\mathbf{r} = \iint(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\,dS = \frac{11}{\sqrt{29}}\iint dS = \frac{11\pi}{\sqrt{29}} $$ EDIT: whoops, forgot the normalization on $\hat{\mathbf{n}}$.