find an expression for a unit vector normal to the surface\n\n$x = 6\\cos(\\theta)\\sin(\\phi),y =…

Explanation:

Step1: Find the partial - derivatives of the position vector

Let $\vec{r}(\theta,\phi)=x(\theta,\phi)\vec{i}+y(\theta,\phi)\vec{j}+z(\theta,\phi)\vec{k}$, where $x = 6\cos(\theta)\sin(\phi)$, $y = 5\sin(\theta)\sin(\phi)$, $z=\cos(\phi)$. First, find $\vec{r}{\theta}=\frac{\partial\vec{r}}{\partial\theta}$: $\vec{r}{\theta}=- 6\sin(\theta)\sin(\phi)\vec{i}+5\cos(\theta)\sin(\phi)\vec{j}+0\vec{k}$ Second, find $\vec{r}{\phi}=\frac{\partial\vec{r}}{\partial\phi}$: $\vec{r}{\phi}=6\cos(\theta)\cos(\phi)\vec{i}+5\sin(\theta)\cos(\phi)\vec{j}-\sin(\phi)\vec{k}$

Step2: Calculate the cross - product $\vec{r}{\theta}\times\vec{r}{\phi}$

[ \begin{align*} \vec{r}{\theta}\times\vec{r}{\phi}&=\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\ -6\sin(\theta)\sin(\phi)&5\cos(\theta)\sin(\phi)&0\ 6\cos(\theta)\cos(\phi)&5\sin(\theta)\cos(\phi)&-\sin(\phi) \end{vmatrix}\ &=\vec{i}\left(-5\cos(\theta)\sin^{2}(\phi)-0\right)-\vec{j}\left(6\sin(\theta)\sin^{2}(\phi)-0\right)+\vec{k}\left(- 30\sin^{2}(\theta)\sin(\phi)\cos(\phi)-30\cos^{2}(\theta)\sin(\phi)\cos(\phi)\right)\ &=-5\cos(\theta)\sin^{2}(\phi)\vec{i}-6\sin(\theta)\sin^{2}(\phi)\vec{j}-30\sin(\phi)\cos(\phi)\vec{k} \end{align*} ]

Step3: Find the magnitude of $\vec{r}{\theta}\times\vec{r}{\phi}$

[ \begin{align*} \left|\vec{r}{\theta}\times\vec{r}{\phi}\right|&=\sqrt{(-5\cos(\theta)\sin^{2}(\phi))^{2}+(-6\sin(\theta)\sin^{2}(\phi))^{2}+(-30\sin(\phi)\cos(\phi))^{2}}\ &=\sqrt{25\cos^{2}(\theta)\sin^{4}(\phi)+36\sin^{2}(\theta)\sin^{4}(\phi) + 900\sin^{2}(\phi)\cos^{2}(\phi)}\ &=\sqrt{(25\cos^{2}(\theta)+36\sin^{2}(\theta))\sin^{4}(\phi)+900\sin^{2}(\phi)\cos^{2}(\phi)}\ &=\sin(\phi)\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)} \end{align*} ]

Step4: Calculate the unit normal vector components

The unit normal vector $\vec{n}=\frac{\vec{r}{\theta}\times\vec{r}{\phi}}{\left|\vec{r}{\theta}\times\vec{r}{\phi}\right|}$. $n_{x}(\theta,\phi)=\frac{-5\cos(\theta)\sin^{2}(\phi)}{\sin(\phi)\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}=\frac{-5\cos(\theta)\sin(\phi)}{\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}$ $n_{y}(\theta,\phi)=\frac{-6\sin(\theta)\sin^{2}(\phi)}{\sin(\phi)\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}=\frac{-6\sin(\theta)\sin(\phi)}{\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}$ $n_{z}(\theta,\phi)=\frac{-30\sin(\phi)\cos(\phi)}{\sin(\phi)\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}=\frac{-30\cos(\phi)}{\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}$

Answer:

$n_{x}(\theta,\phi)=\frac{-5\cos(\theta)\sin(\phi)}{\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}$ $n_{y}(\theta,\phi)=\frac{-6\sin(\theta)\sin(\phi)}{\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}$ $n_{z}(\theta,\phi)=\frac{-30\cos(\phi)}{\sqrt{(25 + 11\sin^{2}(\theta))\sin^{2}(\phi)+900\cos^{2}(\phi)}}$