evaluate \n\n$$\n\\iint_{s} z d s\n$$\nwhere s is the upper half of a sphere of radius 2

Explanation:

Step1: Parametrize the surface

The upper - half of the sphere of radius (a = 2) can be parametrized as (\vec{r}(\theta,\varphi)=\left\langle2\sin\varphi\cos\theta,2\sin\varphi\sin\theta,2\cos\varphi\right\rangle), where (0\leqslant\theta\leqslant2\pi) and (0\leqslant\varphi\leqslant\frac{\pi}{2}).

The formula for the surface element (dS=\left|\vec{r}{\theta}\times\vec{r}{\varphi}\right|d\theta d\varphi).

First, find the partial derivatives:

(\vec{r}_{\theta}=\left\langle - 2\sin\varphi\sin\theta,2\sin\varphi\cos\theta,0\right\rangle)

(\vec{r}_{\varphi}=\left\langle2\cos\varphi\cos\theta,2\cos\varphi\sin\theta,- 2\sin\varphi\right\rangle)

Then, (\vec{r}{\theta}\times\vec{r}{\varphi}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\-2\sin\varphi\sin\theta&2\sin\varphi\cos\theta&0\2\cos\varphi\cos\theta&2\cos\varphi\sin\theta&-2\sin\varphi\end{array}\right|)

(=\vec{i}\left(-4\sin^{2}\varphi\cos\theta\right)-\vec{j}\left(4\sin^{2}\varphi\sin\theta\right)+\vec{k}\left(- 4\sin\varphi\cos\varphi\sin^{2}\theta-4\sin\varphi\cos\varphi\cos^{2}\theta\right))

(=\left\langle-4\sin^{2}\varphi\cos\theta,-4\sin^{2}\varphi\sin\theta,-4\sin\varphi\cos\varphi\right\rangle)

(\left|\vec{r}{\theta}\times\vec{r}{\varphi}\right|=\sqrt{16\sin^{4}\varphi\cos^{2}\theta + 16\sin^{4}\varphi\sin^{2}\theta+16\sin^{2}\varphi\cos^{2}\varphi})

Since (\sin^{2}\theta+\cos^{2}\theta = 1), we have (\left|\vec{r}{\theta}\times\vec{r}{\varphi}\right| = 4\sin\varphi)

Also, (z = 2\cos\varphi)

Step2: Set up the double - integral

The surface integral (\iint_{S}z~dS=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}(2\cos\varphi)\times(4\sin\varphi)d\varphi d\theta)

First, integrate with respect to (\varphi):

Let (u=\sin\varphi), then (du=\cos\varphi d\varphi). When (\varphi = 0), (u = 0); when (\varphi=\frac{\pi}{2}), (u = 1)

(\int_{0}^{\frac{\pi}{2}}8\sin\varphi\cos\varphi d\varphi=8\int_{0}^{1}u~du=8\times\frac{u^{2}}{2}\big|_{0}^{1}=4)

Then integrate with respect to (\theta):

(\int_{0}^{2\pi}4d\theta=4\theta\big|_{0}^{2\pi}=8\pi)

Answer:

(8\pi)