91Ó°ÊÓ

To solve a line integral over a line segment, we first need to parametrize the path the integral follows. This involves creating a function that captures the transition from one point to another. For the line segment connecting the points

• \( P_0 = (1, 1, 1) \)
• \( P_1 = (3, 2, 0) \)

we determine a parametric representation. This is done by introducing a parameter, usually \( t \), which ranges from 0 to 1, representing movement along the line segment. The parameterization is given by:
\[ \mathbf{r}(t) = (1 + 2t, 1 + t, 1 - t) \]Each component in this vector equation corresponds to a linear interpolation between the start and end point coordinates:

• The x-component moves from 1 to 3, so \( x = 1 + 2t \).
• The y-component moves from 1 to 2, so \( y = 1 + t \).
• The z-component decreases from 1 to 0, so \( z = 1 - t \).

We use this parameterization to substitute variables in the line integral later.

In the next step of evaluating a line integral, we express the differential elements \( dx, dy, dz \) in terms of our parameterization. This conversion is necessary for performing integration with respect to the parameter \( t \).To find the differential changes:

• For \( dx \), differentiate \(1 + 2t\) with respect to \( t \), resulting in \( dx = 2 \, dt \).
• For \( dy \), differentiate \(1 + t\) with respect to \( t \), resulting in \( dy = dt \).
• For \( dz \), differentiate \(1 - t\) with respect to \( t \), giving \( dz = -dt \).

These differentials substitute into the line integral, transforming it from a function of \( x, y, z \) to a function of \( t \). This step simplifies integration by aligning it with the parameterization, making the integrand easier to handle.

The final phase involves substituting our parameterized variables and differentials into the original line integral expression, transforming it into a single-variable integral over \( t \).The line integral we need to evaluate is:\[\int_{C} yz \, dx + xz \, dy + xy \, dz\]Substituting the parameterized forms of \( x, y, z \) and their differentials gives:

• For \( yz \, dx \): Replace \( y \) with \( 1 + t \), \( z \) with \( 1 - t \), and \( dx \) with \( 2 \, dt \).
• For \( xz \, dy \): Replace \( x \) with \( 1 + 2t \), \( z \) with \( 1 - t \), and \( dy \) with \( dt \).
• For \( xy \, dz \): Replace \( x \) with \( 1 + 2t \), \( y \) with \( 1 + t \), and \( dz \) with \( -dt \).

The rewritten line integral appears as:\[\int_{0}^{1} ((1 + t)(1 - t) \cdot 2 \, dt) + ((1 + 2t)(1 - t) \cdot dt) + ((1 + t)(1 + 2t) \cdot -dt)\]Simplifying this expression inside the integral combines all terms into one manageable function. Finally, integrate this simplified function over the interval from 0 to 1, resulting in the line integral's value. Understand that each term corresponds to a physical or geometrical component of the problem described by the function surfaces intersecting along this line segment.