Question from my engineering mathematics course:
Let $C$ be the vertical circle given by $y = a \sin t$ and $z = a \cos t$ with $x = 0$. Use Stoke's Theorem to evaluate $\displaystyle \int_C(x^2,y,0)\cdot \mathrm d \mathbf r$. Check your answer by calculating the integral directly.
Answer from solutions:
It is clear that the unit normal vector to $D$ is $\mathbf n = \pm \mathbf i$. By inspecting the diagram we see that the $\mathbf n$ must be $-\mathbf i$ to ensure that $S$ stays on the left as we walk around $C$ in the direction of increasing $t$. We also have $\mathrm dA = \mathrm dy \, \mathrm dz = s \,\mathrm ds \, \mathrm dt$, so
$$\operatorname{curl}(\mathbf u) \cdot \mathrm d\mathbf A=(-1,0,-x^2)\cdot (-\mathbf{i}) s \, \mathrm ds \, \mathrm dt = s \, \mathrm ds \, \mathrm dt$$
$$\int_S \operatorname{curl}(\mathbf u) \cdot \mathrm d\mathbf A=\int_{t = 0}^{2\pi}\int_{s=0}^a s \, \mathrm ds \, \mathrm dt = \int_{t=0}^{2\pi}\frac 1 2 a^2 \, \mathrm dt = \pi a^2$$
... Full Answer
I don't understand how the normal ends up pointing in the -'ve x axis direction.
My understanding says that using the right hand rule, you point your first finger in the direction of the first vector (so in this example, the y axis), second finger points in the direction of the 2nd vector (so the z axis) and then thumb points towards direction of normal (which gives POSITIVE x).
Am I understanding this incorrectly?