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 \(t\):

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

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

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 \(d\mathbf{r}\):

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

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

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

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 \(-\pi\).