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

Line Integral Calculation

The problem asks us to calculate the line integral of the vector function $\mathbf{F}=yz\hat{i}$ along a specified path. The path is a circle defined by $x^2+y^2=1$ at $z=1$, traversed in the counter-clockwise direction.

Understanding the Path

The equation $x^2+y^2=1$ represents a circle of radius 1 centered at the origin in the xy-plane. Since the integration is performed at a constant $z=1$, the path is a unit circle lying on the plane $z=1$.

Parameterizing the Path

To evaluate the line integral ${\oint }_C\mathbf{F}\cdot d\mathbf{r}$, we first parameterize the circular path. A standard parameterization for a unit circle in the counter-clockwise direction is:

- $x(t)=\cos (t)$
- $y(t)=\sin (t)$
- $z(t)=1$

where the parameter $t$ ranges from $0$ to $2\pi$ to complete one full circle.

**Calculating $d\mathbf{r}$ **

Next, we find the differential displacement vector $d\mathbf{r}$ by differentiating the parametric equations with respect to $t$:

- $dx=-\sin (t)dt$
- $dy=\cos (t)dt$
- $dz=0dt$

Therefore, $d\mathbf{r}=dx\hat{i}+dy\hat{j}+dz\hat{k}=(-\sin (t)dt)\hat{i}+(\cos (t)dt)\hat{j}+(0dt)\hat{k}$.

Substituting into the Vector Field

Now, substitute the parameterized values of $x,y,z$ into the vector function $\mathbf{F}=yz\hat{i}$:

$\mathbf{F}=(\sin (t))(1)\hat{i}=\sin (t)\hat{i}$

**Calculating the Dot Product $\mathbf{F}\cdot d\mathbf{r}$ **

Compute the dot product of $\mathbf{F}$ and $d\mathbf{r}$:

$\mathbf{F}\cdot d\mathbf{r}=(\sin (t)\hat{i})\cdot ((-\sin (t)dt)\hat{i}+(\cos (t)dt)\hat{j}+(0dt)\hat{k})$

$\mathbf{F}\cdot d\mathbf{r}=\sin (t)\cdot (-\sin (t)dt)$

$\mathbf{F}\cdot d\mathbf{r}=-{\sin }^2(t)dt$

Evaluating the Line Integral

The line integral is the integral of $\mathbf{F}\cdot d\mathbf{r}$ over the range of the parameter $t$:

${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\int }_0^{2\pi }-{\sin }^2(t)dt$

To solve this integral, we use the trigonometric identity ${\sin }^2(t)=\frac{1-\cos (2t)}{2}$:

${\int }_0^{2\pi }-\left(\frac{1-\cos (2t)}{2}\right)dt=-\frac{1}{2}{\int }_0^{2\pi }(1-\cos (2t))dt$

Now, integrate with respect to $t$:

$=-\frac{1}{2}{\left[t-\frac{\sin (2t)}{2}\right]}_0^{2\pi }$

Evaluate the definite integral:

$=-\frac{1}{2}\left[\left(2\pi -\frac{\sin (4\pi )}{2}\right)-\left(0-\frac{\sin (0)}{2}\right)\right]$

$=-\frac{1}{2}\left[(2\pi -0)-(0-0)\right]$

$=-\frac{1}{2}[2\pi ]$

$=-\pi$

Final Result

The value of the line integral is $-\pi$.