I am asked to find $\int_{C} \textbf{F} d\textbf{r}$ using Stokes' Theorem given $\textbf{F} = (z^2, 2x, y^2)$ and $S: z = 1-x^2-y^2$ for $z \geq 0$.
I am proceeding that calculation in two ways:
Using $\displaystyle \int_{C} \textbf{F} d\textbf{r}$
$$\int_{C} \textbf{F} d\textbf{r}=\int_{C} \textbf{F}(\textbf{r}(t)) \cdot \textbf{r'}(t) dt= \begin{bmatrix} x= \cos t\\ y= \sin t \end{bmatrix} = \int_{0}^{2\pi} (0,2\cos t, \sin^2 t) \cdot (-\sin t, \cos t, 0) dt = \int_{0}^{2\pi} 2\cos^2 t dt = \int_{0}^{2\pi} 1+\cos 2t dt =2\pi$$
Which is the correct answer (according to the textbook)
Using $\displaystyle \iint_S curl \ \textbf{F} \cdot d\textbf{S}$
$$P=(x,y,1-x^2-y^2) \quad \frac{\partial P}{\partial x}=(1,0,-2x) \quad \frac{\partial P}{\partial y}=(0,1,-2y)$$
$$\frac{\partial P}{\partial x} \times \frac{\partial P}{\partial y} = (1,0,-2x) \times (0,1,-2y)=(2x,2y,1)$$
And exactly on this step I am not sure what I should do: to evaluate the $curl \textbf{F}$ should I use
$$curl \ \textbf{F}= \begin{bmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ z^2 & 2x & y^2 \end{bmatrix}$$
or
$$curl \ \textbf{F}= \begin{bmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ (1-x^2-y^2)^2 & 2x & y^2 \end{bmatrix}$$
Thank you.