Question : Find curl of r^n vec(r)
Solution :
We shall Find curl of r^n vec(r) We know that \[\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partialy}+\hat{k}\frac{\partial}{\partialz}\] And \[r^n\vec{r}\ =r^n\left(x\hat{i}+y\hat{j}+z\hat{k}\right)=r^nx\hat{i}+r^ny\hat{j}+r^nz\hat{k}\] Therefore curl of r^n vec(r) is \[\nabla\times\left(r^n\vec{r}\right)=\left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\r^nx&\ r^ny&\ r^nz\\\end{matrix}\right|\] \[=\hat{i}\left(\frac{\partial\left(\ r^nz\right)}{\partial y}-\frac{\partial\left(r^ny\right)}{\partial z}\right)-\hat{j}\left(\frac{\partial\left(\ r^nz\right)\ }{\partial x}-\frac{\partial\left(r^nx\right)}{\partial z}\right)+\hat{k}\left(\frac{\partial\left(r^ny\right)}{\partial x}-\frac{\partial\left(r^nx\right)}{\partial y}\right)\] \[=\hat{i}\left(znr^{n-1}\frac{\partial r}{\partial y}-ynr^{n-1}\frac{\partial r}{\partial z}\right)-\hat{j}\left(znr^{n-1}\frac{\partial r}{\partial x}-xnr^{n-1}\frac{\partial r}{\partial z}\right)+\hat{k}\left(ynr^{n-1}\frac{\partial r}{\partial x}-xnr^{n-1}\frac{\partial r}{\partial y}\right)\] We know that partial derivative \[\frac{\partial r}{\partial x}=\frac{x}{r},\frac{\partial r}{\partial y}=\frac{y}{r}\ ,\ \frac{\partial r}{\partial z}=\frac{z}{r}\] substituting these values we get \[\nabla\times\left(r^n\vec{r}\right)=nr^{n-2}\left\{\hat{i}\left(zy-yz\right)-\hat{j}\left(zx-xz\right)+\hat{k}\left(yx-xy\right)\right\}\] \[\nabla\times\left(r^n\vec{r}\right)=nr^{n-2}\left\{\hat{i}0-\hat{j}0+\hat{k}0\right\}=0\] \[\nabla\times\left(r^n\vec{r}\right)=0\] Thus we found that curl of r^n vec(r) is zero.