I'm trying to solve this question:
Let $\underline u \colon \mathbb R^3 \to \mathbb R^3$ be a vector field defined as $$ \underline u(x,y,z) = (x^3+2y+z,\,y^3+2x+z,\,1/2(x^2+y^2)+x+y)\,. $$ Let $K$ be the curve of intersection of the surfaces $V$ and $S$, where $V: x+y-z=1$ and $$ S: \begin{cases} z = 1-x^2-y^2\,, \\ z \geq 0\,. \end{cases} $$
Calculate $$ \int_K \underline u \bullet\underline t \,ds\,, $$ where $K$ is along the path $(1,0,0)$ to $(0,1,0)$.
The answer appears to be $$ \int_K \underline u \bullet\underline t \,ds = 0 $$
We've been learning Green and Gauss' theorems, as well as Stokes' theorem in class. I figured Stokes theorem is best applicable here, as we don't want to calculate the flux, but rather the work done when moving along this curve.
I first calculated the curl and divergence; \begin{align*} \text{div}\underline u &= 3x^2 + 3y^2 \\ \text{curl}\underline u &= (y,-x,0) \end{align*}
We now know, by Stokes' theorem $$ \iint_S \text{curl} \underline u \bullet \underline n \,d\sigma = \oint_{K'} \underline u \bullet \underline t \,ds\,, $$ if $K'$ is the complete inclined ellipse (where $K$ is a small part of) and $S$ is the surface of this ellipse.
I have found parametrizations of $K'$ (and $S$), but these are not pretty; \begin{align*} k(t) &= \left(-1/2+\sqrt{5/2}\cos t, -1/2+\sqrt{5/2}\sin t, -2+\sqrt{5/2}(\cos t + \sin t)\right)\,, \\ s(r,t) &= \left(s\left[-1/2+\sqrt{5/2}\cos t\right], s\left[-1/2+\sqrt{5/2}\sin t\right], -1+s\left[-1+\sqrt{5/2}(\cos t + \sin t)\right]\right)\,, \end{align*} where $t \in [0,2\pi]$ and $s \in [0,1]$.
Now I could try to work out this integral: $$ \iint_S \text{curl}\underline v \bullet \underline n \,d\sigma = \int_0^{2\pi}\int_0^1 \text{curl}(v(s(r,t))\bullet \left( \frac{\partial k}{\partial r} \times \frac{\partial k}{\partial t} \right)\,dr dt\,, $$ but I don't think thats really doable, plus I would only know the work done for the the full ellipse ($K'$) and not for the small part ($K$).
I was also thinking about the fact that the normal on $S$ is the same for every point. I could use the vector $\underline n = (-1,-1,2)$, and normalize $\underline n' = \frac{1}{\sqrt{5}}(-1,-1,2)$, and write $$ \iint_S \text{curl}\underline v \bullet \underline n \,d\sigma = \iint_S (y,-x,0)\bullet \frac{1}{\sqrt{5}}(-1,-1,2) \,d\sigma\,, $$ yet I would probably still need to use the parametrization.
To tackle the problem directly (without using any of the theorems), I would need to find the start and end points of $k(t)$, which is also difficult by hand.
What is the best way to approach this problem?
Hint. Use Stokes theorem $$\oint_{K\cup K'} \underline u \bullet \underline t \,ds=\iint_S \text{curl} \underline u \bullet \underline n \,d\sigma$$ where $S$ is the surface given by the the part of the plane $x+y-z=1$ inside the closed curve $K\cup K'$ where $K'$ is the segment $[0,1]\ni t\to (t,1-t,0)$. Note that the normal to the plane is $\underline n=(1,1,-1)/\sqrt{3}$. Hence $$\frac{1}{\sqrt{3}}\iint_S (y,-x,0)\bullet (1,1,-1) \,d\sigma =\frac{1}{\sqrt{3}}\iint_S y \,d\sigma-\frac{1}{\sqrt{3}}\iint_S x \,d\sigma=0$$ because $S$ is symmetric with respect to the plane $y=x$.
Finally the integral along $K'$ is easy to evaluate: $$\int_{K'}\underline u \bullet \underline t\, ds =\int_0^1 [(t^3+2(1-t))\cdot(1)+((1-t)^3+2t)\cdot(-1)]dt=0.$$