Let $\mathbf{F} = (x^2 - z, xy - x, xz + x)$ be a vector field, and let $S$ be the unit sphere in $\mathbb{R}^3$.
Determine the simple, closed curve $\gamma$ (orientation does not need to be determined) lying on the surface $S$ that maximizes the line integral $$\int_\gamma \mathbf{F} \cdot d\mathbf{r}$$
Answer:
Since the curve is simple, it must enclose an area $Z$ on the unit sphere, and Stokes' theorem states that
$$\int_\gamma \mathbf{F} \cdot d\mathbf{r} = \iint_Z \text{rot} \mathbf{F} \cdot \mathbf{n} \, dS = \iint_Z [0, -2 - z, -1 + y] \cdot [x, y, z] \, dS = \iint_Z (-2y - z) \, dS$$
This integral is maximized when $(x, y, z)$ traverses all points where $-2y - z > 0$. The boundary $\gamma$ is therefore given by the intersection of the sphere with the plane $-2y - z = 0$. Substituting $z = -2y$ into the sphere equation, we obtain $$1 = x^2 + y^2 + (-2y)^2 = x^2 + 5y^2$$ Thus, the curve $\gamma$ is a great circle parametrized by $$\mathbf{r}(t) = [\cos t, \frac{1}{\sqrt{5}} \sin t, -\frac{2}{\sqrt{5}} \sin t]$$
So I don't understand the part where they are saying $-2y - z > 0$ and then they go and make $-2y - z = 0 $ why is that?