Vector calculus: Outer product and surface integral

Question:
A curved surface, $S$, is given as
$\vec{r}=\cos u \sin v \vec{i} + \sin u \sin v \vec{j} + \cos v \vec{k}$
$(0\leqq u \leqq 2\pi, 0\leqq v \leqq \pi)$
(1) Find $\frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}$.
(2) Find $\iint_S \left| \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}\right| dudv$.

Answer:
(1) Take the partial derivative.
$$\begin{eqnarray} \frac{\partial\vec{r}}{\partial u} &=& -\sin u \sin v \vec{i} + \cos u \sin v \vec{j}  \\ \frac{\partial\vec{r}}{\partial v} &=& \cos u \cos v \vec{i} + \sin u \cos v \vec{j} -\sin v \vec{k} \end{eqnarray}$$
Therefore,
$$\frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v} = \left| \begin{array}{c} \vec{i}  &  \vec{j}  &  \vec{k}   \\ \left(\frac{\partial\vec{r}}{\partial u}\right)_i & \left(\frac{\partial\vec{r}}{\partial u}\right)_j & \left(\frac{\partial\vec{r}}{\partial u}\right)_k    \\ \left(\frac{\partial\vec{r}}{\partial v}\right)_i & \left(\frac{\partial\vec{r}}{\partial v}\right)_j & \left(\frac{\partial\vec{r}}{\partial v}\right)_k     \\ \end{array} \right|   \\ = \left| \begin{array}{c} \vec{i}  &  \vec{j}  &  \vec{k}   \\ -\sin u \sin v & \cos u \sin v & 0    \\ \cos u \cos v & \sin u \cos v & -\sin v     \\ \end{array} \right|   \\ = \left| \begin{array}{c} \cos u \sin v & 0    \\ \sin u \cos v & -\sin v     \\ \end{array} \right|\vec{i} - \left| \begin{array}{c} -\sin u \sin v & 0    \\ \cos u \cos v & -\sin v     \\ \end{array} \right| \vec{j} + \left| \begin{array}{c} -\sin u \sin v & \cos u \sin v     \\ \cos u \cos v & \sin u \cos v      \\ \end{array} \right|\vec{k}    \\ =-\cos u \sin^2 v\vec{i}-\sin u \sin^2 v\vec{j}-\sin v \cos v(\sin^2 u +\cos^2 u)\vec{k}  \\ =-\cos u \sin^2 v\vec{i}-\sin u \sin^2 v\vec{j}-\sin v \cos v \vec{k}$$

(2) Calculate $\left| \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}\right|$.
$$\begin{eqnarray*} \left| \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}\right| &=& \sqrt{|-\cos u \sin^2 v\vec{i}-\sin u \sin^2 v\vec{j}-\sin v \cos v \vec{k}|^2}  \\ &=& \sqrt{\cos^2 u \sin^4 v+\sin u^2 \sin^4 v+\sin^2 v \cos^2 v}   \\ &=& \sqrt{\sin^4 v(\cos^2 u+\sin u^2)+\sin^2 v \cos^2 v}   \\ &=& \sqrt{\sin^2 v(\sin^2 v+\cos^2 v)}   \\ &=& |\sin v| \end{eqnarray*}$$
Therefore, we have the integral.
$$\begin{eqnarray*} \int^{\pi}_0 |\sin v| dv \int^{2\pi}_0 du &=& [-\cos v]^{\pi}_0 \cdot 2\pi  \\      &=& [-\cos \pi -(-\cos 0)] \cdot 2\pi   \\      &=& [ 1 + 1] \cdot 2\pi  \\      &=& 4\pi \end{eqnarray*}$$

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