Divergence & Curl Engineering Maths, Btech first year

Divergence & Curl

Divergence

If $\overrightarrow{F}⟶$ Vector Point Function + Continuous + Differentiable then divergence of $\overrightarrow{F}$ is $$div\overrightarrow{F}=\mathrm{\nabla }.\overrightarrow{F}$$ i.e. dot product between $\mathrm{\nabla }$ and $\overrightarrow{F}$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z}).(F_1\hat{i}+F_2\hat{j}+F_3\hat{k})$$

$$\boxed{{\displaystyle div\overrightarrow{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}}}$$

Note : The function is vector but its divergence is scalar.

Solenoidal Vector :

A vector is said to be Solenoidal if $div\overrightarrow{F}=0$ $$\boxed{{\displaystyle \mathrm{\nabla }.\overrightarrow{F}=0}}$$

Sample Problem 1 :

Evaluate $div(3x^2\hat{i}+5x{y}^2\hat{j}+xy{z}^3\hat{k})$ at point (1,2,3).

Solution

Let $f=3x^2\hat{i}+5x{y}^2\hat{j}+xy{z}^3\hat{k}$ $$div(f)=\mathrm{\nabla }.f$$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z}).(3x^2\hat{i}+5x{y}^2\hat{j}+xy{z}^3\hat{k})$$ $$=6x+10xy+3x{y}^2$$ At point (1,2,3) $$div(f)=6(1)+10(1)(2)+3(1)(2)(3{)}^2$$ $$=\boxed{{\displaystyle 80}}$$

Curl of a Vector

If $\overrightarrow{F}⟶$ Vector Point Function + Continuous + Differentiable
then $curl(\overrightarrow{F})=\mathrm{\nabla }\times \overrightarrow{F}=$ $$(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})\times (F_1\hat{i}+F_2\hat{j}+F_3\hat{k})$$

$$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \\ F_1& F_2& F_3\end{array}\right|$$

Note : The function is vector and its curl is also a vector.

Irrotational Vector Field :

A vector field is said to be irrotational if $curl(\overrightarrow{F})=0$ $$\boxed{{\displaystyle \mathrm{\nabla }\times \overrightarrow{F}=0}}$$

Sample Problem 2 :

Find curl of $\overrightarrow{F}=xy\hat{i}+y^2\hat{j}+zx\hat{k}$ at point (-2,4,1).

Solution :

$$curl(\overrightarrow{F})=\mathrm{\nabla }\times \overrightarrow{F}$$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})\times (xy\hat{i}+y^2\hat{j}+zx\hat{k})$$ $$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \\ xy& y^2& zx\end{array}\right|$$ $$=\hat{i}[\frac{\partial }{\partial y}(zx)-\frac{\partial }{\partial z}(y2)]$$ $$-\hat{j}[\frac{\partial }{\partial x}(zx)-\frac{\partial }{\partial z}(xy)]$$ $$+\hat{k}[\frac{\partial }{\partial x}(y^2)-\frac{\partial }{\partial y}(xy)]$$ $$=\hat{i}[0-0]-\hat{j}[z-0]+\hat{k}[0-x]$$ $$=0\hat{i}-z\hat{j}-x\hat{k}$$ At point (-2,4,1) $$curl(\overrightarrow{F})=-(1)\hat{j}-(-2)\hat{k}$$ $$\Rightarrow \boxed{{\displaystyle curl(\overrightarrow{F})=-\hat{j}+2\hat{k}}}$$

Sample Problem 3 :

Find $div(\overrightarrow{F})$ & $curl(\overrightarrow{F})$ where $\overrightarrow{F}=grad(x^3+y^3+z^3-3xyz)$

Solution :

Here we haven't been given the vector function directly, instead we have to first find gradient and then evaluate divergence and curl. $$\overrightarrow{F}=grad(x^3+y^3+z^3-3xyz)$$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})(x^3+y^3+z^3-3xyz)$$ $$\Rightarrow \overrightarrow{F}=(3x^2-3yz)\hat{i}+(3y^2-3xz)\hat{j}+(3z^2-3xy)\hat{k}$$ $$\text{Now }div(\overrightarrow{F})=\mathrm{\nabla }.\overrightarrow{F}$$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})[(3x^2-3yz)\hat{i}+(3y^2-3xz)\hat{j}+(3z^2-3xy)\hat{k}]$$ $$\Rightarrow div(\overrightarrow{F})=6x\hat{i}+6y\hat{j}+6z\hat{k}$$ $$\text{Now }curl(\overrightarrow{F})=\mathrm{\nabla }\times \overrightarrow{F}$$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})\times [(3x^2-3yz)\hat{i}+(3y^2-3xz)\hat{j}+(3z^2-3xy)\hat{k}]$$ $$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \\ 3x^2-3yz& 3y^2-3xz& 3z^2-3xy\end{array}\right|$$ $$=\hat{i}[\frac{\partial }{\partial y}(3z^2-3xy)-\frac{\partial }{\partial z}(3y^2-3xz)]$$ $$-\hat{j}[\frac{\partial }{\partial x}(3z^2-3xy)-\frac{\partial }{\partial z}(3x^2-3yz)]$$ $$+\hat{k}[\frac{\partial }{\partial x}(3y^2-3xz)-\frac{\partial }{\partial y}(3x^2-3yz)]$$ $$=\hat{i}[(0-3x)-(0-3x)]-\hat{j}[(0-3y)-(0-3y)]+\hat{k}[(0-3z)-(0-3z)]$$ $$\Rightarrow curl(\overrightarrow{F})=0\hat{i}+0\hat{j}+0\hat{k}=0$$

Sample Problem 4

If $\overrightarrow{R}=x\hat{i}+y\hat{j}+z\hat{k}$ then prove that $div(r^n\overrightarrow{R})=(n+3)r^n$ and find the value of $n$ if $r^n\overrightarrow{R}$ is a solenoidal vector.

Solution

$$\overrightarrow{R}=x\hat{i}+y\hat{j}+z\hat{k}$$ $$r=|\overrightarrow{R}|=\sqrt{x^2+y^2+z^2}$$ L.H.S. = $div(r^n\overrightarrow{R})=\mathrm{\nabla }{r}^n\overrightarrow{R}$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})[r^n(x\hat{i}+y\hat{j}+z\hat{k})]$$ $$=\frac{\partial }{\partial x}(x{r}^n)+\frac{\partial }{\partial y}(y{r}^n)+\frac{\partial }{\partial z}(z{r}^n)$$ $$=(r^n+nx{r}^{n-1}\frac{\partial r}{\partial x})+(r^n+ny{r}^{n-1}\frac{\partial r}{\partial y})+(r^n+nz{r}^{n-1}\frac{\partial r}{\partial z})$$ $$\text{Now }\frac{\partial r}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2+z^2}}=\frac{x}{r}$$ $$\frac{\partial r}{\partial y}=\frac{y}{r},\frac{\partial r}{\partial z}=\frac{z}{r}$$ Using these values $$\mathrm{\nabla }{r}^n\overrightarrow{R}=3r^n+nx{r}^{n-1}\times \frac{x}{r}+ny{r}^{n-1}\times \frac{y}{r}+nz{r}^{n-1}\times \frac{z}{r}$$ $$=3r^n+n{r}^{n-2}(x^2+y^2+z^2)$$ $$=3r^n+n{r}^{n-2}(r^2)$$ $$=(n+3)r^n$$ = R.H.S., Hence proved :)
Since $r^n\overrightarrow{R}$ is a solenoidal vector $$\Rightarrow div(r^n\overrightarrow{R})=0$$ $$\Rightarrow (n+3)r^n=0$$ $$\text{As }{r}^n\ne 0$$ $$\Rightarrow n+3=0$$ $$\Rightarrow \boxed{{\displaystyle n=-3}}$$

Sample Problem 5

If $\overrightarrow{R}=x\hat{i}+y\hat{j}+z\hat{k}$ then prove that $curl(r^n\overrightarrow{R})=0$ and find value of $n$ if $r^n\overrightarrow{R}$ is irrotational vector.

Solution

$$\overrightarrow{R}=x\hat{i}+y\hat{j}+z\hat{k}$$ $$r=|\overrightarrow{R}|=\sqrt{x^2+y^2+z^2}$$ L.H.S. = $curl(r^n\overrightarrow{R})=\mathrm{\nabla }{r}^n\overrightarrow{R}$ $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})\times [r^n(x\hat{i}+y\hat{j}+z\hat{k})]$$ $$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \\ x{r}^n& y{r}^n& z{r}^n\end{array}\right|$$ $$=\hat{i}[\frac{\partial }{\partial y}(z{r}^n)-\frac{\partial }{\partial x}(y{r}^n)]$$ $$-\hat{j}[\frac{\partial }{\partial x}(z{r}^n)-\frac{\partial }{\partial z}(x{r}^n)]$$ $$+\hat{k}[\frac{\partial }{\partial x}(y{r}^n)-\frac{\partial }{\partial y}(x{r}^n)]$$ $$=\hat{i}[zn{r}^{n-1}\frac{\partial r}{\partial y}-yn{r}^{n-1}\frac{\partial r}{\partial z}]$$ $$-\hat{j}[zn{r}^{n-1}\frac{\partial r}{\partial x}-xn{r}^{n-1}\frac{\partial r}{\partial z}]$$ $$+\hat{k}[yn{r}^{n-1}\frac{\partial r}{\partial x}-xn{r}^{n-1}\frac{\partial r}{\partial y}]$$ As we had calculated in previous problems $$\frac{\partial r}{\partial x}=\frac{x}{r},\frac{\partial r}{\partial y}=\frac{y}{r},\frac{\partial r}{\partial z}=\frac{z}{r}$$ $$=\hat{i}[\cancel{zn{r}^{n-1}\times \frac{y}{r}}-\cancel{yn{r}^{n-1}\times \frac{y}{r}}]$$ $$-\hat{j}[\cancel{zn{r}^{n-1}\times \frac{x}{r}}-\cancel{xn{r}^{n-1}\times \frac{z}{r}}]$$ $$+\hat{k}[\cancel{yn{r}^{n-1}\times \frac{x}{r}}-\cancel{xn{r}^{n-1}\times \frac{y}{r}}]$$ $$\Rightarrow curl(r^n\overrightarrow{R})=0$$ Hence Proved :)
Since $r^n\overrightarrow{R}$ is irrotational $$\Rightarrow curl(r^n\overrightarrow{R})=0$$ which we already proved true, so $n$ can have any value.