The line integral of function $F=yz\hat{i}$, in the counter clock wise direction, along the circle $x^2+y^2=1$ at $z=1$ is

Concept:

Green's theorem:

$\oint \left(\mathrm{M}\mathrm{d}\mathrm{x}+\mathrm{N}\mathrm{d}\mathrm{y}\right)=\int \int \left(\frac{\partial \mathrm{N}}{\partial \mathrm{x}}-\frac{\partial \mathrm{M}}{\partial \mathrm{y}}\right)\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}$

Calculation:

$\begin{array}{l}F=yz\hat{i}\\ \underset{c}{\overset{}{\int }}F.dr=\underset{c}{\overset{}{\int }}\left(yz\hat{i}\right)\left(\hat{i}dx+\hat{j}dy+\hat{k}dz\right)\\ =\underset{c}{\overset{}{\int }}yzdx\\ =\underset{c}{\overset{}{\int }}ydx\left(\therefore z=1given\right)\\ =\underset{c}{\overset{}{\int }}\left(ydx-ody\right)\end{array}$

From Green’s Theorem

$\begin{array}{l}\underset{c}{\overset{}{\oint }}Pdx+Qdy=\int \underset{R}{\overset{}{\int }}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy\\ \therefore \underset{c}{\overset{}{\int }}ydx-0dy=\int \underset{x^2+y^2=1}{\overset{}{\int }}\left(0-1\right)dxdy\\ \left(P=y,Q=0\right)\\ =\int \underset{x^2+y^2=1}{\overset{}{\int }}-dxdy=-\int \underset{x^2+y^2=1}{\overset{}{\int }}dxdy\end{array}$

= - Area of circle (x² + y² = 1)

= - π × (1)²

= - π

$\underset{c}{\overset{}{\int }}\overrightarrow{F}.d\overrightarrow{r}=-\pi$

NOTE - 

Stoke's theorem:

$\underset{C}{\overset{}{\oint }}\overrightarrow{A}.d\overrightarrow{l}=\int \underset{S}{\overset{}{\int }}\left(\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{A}\right).d\overrightarrow{s}=\int \underset{S}{\overset{}{\int }}\overrightarrow{V}.d\overrightarrow{s}$

Gauss Divergence theorem:

$\oint A.ds=\iiint (\mathrm{\nabla }.A)dv$

$\oint \overrightarrow{F.}\hat{n}ds=\iiint (\mathrm{\nabla }.F)dv$