91影视

The given equation of the sphere of radius r and centered at the origin is

$x^2+y^2+z^2=k^2.........\left(1\right)$

$$\begin{gathered} \begin{array}{r}{\mathbf{r}}_u脳{\mathbf{r}}_v=鉄�鈭�k\sin v\sin u,k\sin v\cos u,0鉄�脳鉄�k\cos v\cos u,k\cos v\sin u,鈭�k\sin v鉄�\\ =\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 鈭�k\sin v\sin u& k\sin v\cos u& 0\\ k\cos v\cos u& k\cos v\sin u& 鈭�k\sin v\end{array}\right|\end{array} \\ \begin{array}{r}=\left[\right(k\sin v\cos u\left)\right(鈭�k\sin v)鈭�(0\left)\right(k\cos v\sin u\left)\right]\mathbf{i}\\ 鈭�\left[\right(鈭�k\sin v\sin u\left)\right(鈭�k\sin v)鈭�(0\left)\right(k\cos v\cos u\left)\right]\mathbf{j}\\ +\left[\right(鈭�k\sin v\sin u\left)\right(k\cos v\sin u)鈭�(k\sin v\cos u\left)\right(k\cos v\cos u\left)\right]\mathbf{k}\\ =鈭�k^2{\sin }^{2}v\cos u\mathbf{i}鈭�k^2{\sin }^{2}v\sin u\mathbf{j}鈭�k^2\sin v\cos v\left({\sin }^{2}u+{\cos }^{2}u\right)\mathbf{k}\\ =鈭�k^2{\sin }^{2}v\cos u\mathbf{i}鈭�k^2{\sin }^{2}v\sin u\mathbf{j}鈭�k^2\sin v\cos v\left(1\right)\mathbf{k}\\ =鈭�k^2{\sin }^{2}v\cos u\mathbf{i}鈭�k^2{\sin }^{2}v\sin u\mathbf{j}鈭�k^2\sin v\cos v\mathbf{k}.\end{array} \end{gathered}$$

$\begin{array}{r}鈭�{\mathbf{r}}_u脳{\mathbf{r}}_v鈭�=鈭�鈭�k^2{\sin }^{2}v\cos u\mathbf{i}鈭�k^2{\sin }^{2}v\sin u\mathbf{j}鈭�k^2\sin v\cos v\mathbf{k}鈭�\\ =\sqrt{{\left(鈭�k^2{\sin }^{2}v\cos u\right)}^2+{\left(鈭�k^2{\sin }^{2}v\sin u\right)}^2+{\left(鈭�k^2\sin v\cos v\right)}^2}\\ =\sqrt{k^4{\sin }^{4}v{\cos }^{2}u+k^4{\sin }^{4}v{\sin }^{2}u+k^4{\sin }^{2}v{\cos }^{2}v}\\ =\sqrt{\left({\sin }^{2}v{\cos }^{2}u+{\sin }^{2}v{\sin }^{2}u+{\cos }^{2}v\right)k^4{\sin }^{2}v}\\ \begin{array}{r}=\sqrt{\left(\left({\cos }^{2}u+{\sin }^{2}u\right){\sin }^{2}v+{\cos }^{2}v\right)k^4{\sin }^{2}v}\\ =\sqrt{\left(\left(1\right){\sin }^{2}v+{\cos }^{2}v\right)k^4{\sin }^{2}v}\\ =\sqrt{\left({\sin }^{2}v+{\cos }^{2}v\right)k^4{\sin }^{2}v}\\ =\sqrt{\left(1\right)k^4{\sin }^{2}v}\\ =\sqrt{k^4{\sin }^{2}v}\\ =k^2\sin \end{array}\\ \begin{array}{r}dS=\left(鈭�{\mathbf{r}}_x脳{\mathbf{r}}_y鈭�\right)dA\\ =\left(k^2\sin 鈦�v\right)dA\end{array}\end{array}$