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\(12-18\) (a) Find a function \(f\) such that \(\mathbf{F}=\nabla f\) and \((b)\) use part (a) to evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along the given curve \(C .\) $$\mathbf{F}(x, y, z)=y z \mathbf{i}+x z \mathbf{j}+(x y+2 z) \mathbf{k}$$ \(C\) is the line segment from \((1,0,-2)\) to \((4,6,3)\)

Short Answer

Expert verified
The line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) equals 77.

Step by step solution

01

Verify if \( \mathbf{F} \) is Conservative

For \(\mathbf{F}\) to be conservative, the curl \( abla \times \mathbf{F} \) must be zero. Compute the curl:\[abla \times \mathbf{F} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \yz & xz & (xy + 2z) \\end{vmatrix}\]Calculate the partial derivatives, and confirm that all components of the curl vector are zero, confirming that \( \mathbf{F} \) is conservative.
02

Find the Potential Function \( f \)

To find \( f \) such that \( \mathbf{F} = abla f \), integrate each component of \( \mathbf{F} \) with respect to its variable:1. Integrate \( yz \) with respect to \( x \) to get \( f(x, y, z) = xyz + g(y, z) \).2. Integrate \( xz \) with respect to \( y \): The function \( g(y,z) \) must result in consistency with the other components.3. Integrate \( (xy + 2z) \) with respect to \( z \): Obtain \( f(x, y, z) = xyz + z^2 + C \).Combine all parts into a single function ensuring consistency across each partial integration.
03

Evaluate the Line Integral Using the Potential Function

Since \( \mathbf{F} \) is conservative, the line integral is independent of the path and can be computed using the fundamental theorem for line integrals:\[ \int_{C} \mathbf{F} \cdot d\mathbf{r} = f(4, 6, 3) - f(1, 0, -2) \]Substitute the endpoints into the potential function:1. Calculate \( f(4, 6, 3) = 4 \times 6 \times 3 + 3^2 = 72 + 9 = 81 \).2. Calculate \( f(1, 0, -2) = 1 \times 0 \times (-2) + (-2)^2 = 0 + 4 = 4 \).Subtract the two results to find the value of the line integral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Function
In vector calculus, a potential function is a scalar function whose gradient yields a given vector field. If you have a vector field \( \mathbf{F} = abla f \), then \( f \) is considered the potential function of \( \mathbf{F} \). To find the potential function, you need to identify the function \( f \) such that the gradient \( abla f \) matches the components of \( \mathbf{F} \). This essentially involves integration of the vector field components.

In the exercise, the vector field \( \mathbf{F}(x, y, z) = yz \mathbf{i} + xz \mathbf{j} + (xy + 2z) \mathbf{k} \) needed to be matched with a gradient. Each component of \( \mathbf{F} \) was integrated with respect to its variable. This produced part of the potential function:
  • Integrating \( yz \) with respect to \( x \) resulted in \( xyz + g(y, z) \).
  • Integrating \( xz \) with respect to \( y \) ensured compatibility with other components.
  • Integrating \( (xy + 2z) \) with respect to \( z \) completed the function as \( xyz + z^2 + C \).
The final result, \( f(x, y, z) = xyz + z^2 + C \), is the desired potential function consistent with \( \mathbf{F} = abla f \). The constant \( C \) arises as integration constants, as gradient fields only determine the potential function up to an additive constant.
Line Integral
A line integral, in the context of vector fields, measures the magnitude of a vector field along a curve. This is akin to accumulating tiny bits of the vector field's influence as one moves along the curve
  • The integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) captures the total effect of \( \mathbf{F} \) as it passes along the path \( C \).
  • A line integral takes into account not only the path \( C \) itself but also the inclination of \( \mathbf{F} \) relative to \( C \).
To find a line integral, you traditionally parameterize the path \( C \) and evaluate \( \mathbf{F} \cdot d\mathbf{r} \) along it. However, when \( \mathbf{F} \) is conservative, this calculation can be greatly simplified.

In our exercise, instead of following the traditional path-dependent method, we were able to utilize the potential function to determine the integral across \( C \) by only considering the endpoints, thanks to the next concept: the fundamental theorem for line integrals.
Fundamental Theorem for Line Integrals
The fundamental theorem for line integrals elegantly simplifies the computation of line integrals in conservative vector fields. When a vector field \( \mathbf{F} \) is conservative, this theorem states that the line integral along a path \( C \) can be calculated simply as the difference of the potential function \( f \) between the endpoints of the path:
  • \( \int_{C} \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A) \), where \( A \) and \( B \) are start and end points on \( C \).
This theorem is extremely powerful as it eliminates the need to perform tedious parameterization and integration along the path.

In the exercise, the path was a straight line segment from \( (1,0,-2) \) to \( (4,6,3) \). By applying the fundamental theorem, we computed the potential function at these points:
  • \( f(4,6,3) = 81 \)
  • \( f(1,0,-2) = 4 \)
Taking their difference, \( 81 - 4 = 77 \), provided the value of the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \). This direct method showcases the convenience offered by the theorem in dealing with conservative fields.

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