Question

In Exercises \(17-20\), a conservative vector field \(\vec{F}\) and a curve C are given. 1\. Find a potential function \(f\) for \(\vec{F}\) 2\. Compute curl \(\vec{F}\). 3\. Evaluate \(\int_{C} \vec{F} \cdot d \vec{r}\) directly, i.e., using Key Idea 14.3.1. 4\. Evaluate \(\int_{C} \vec{F} \cdot d \vec{r}\) using the Fundamental Theorem of Line Integrals. $$ \begin{array}{l} \vec{F}=\left\langle 2 x y z, x^{2} z, x^{2} y\right\rangle, c \text { is curve parametrized by } \vec{r}(t)= \\ \langle 2 t+1,3 t-1, t\rangle \text { on } 0 \leq t \leq 2 \end{array} $$

Short answer

The line integral evaluates to 250.

Step-by-step solution

Find the potential function

We have \( \vec{F} = \langle 2xy, x^2z, x^2y \rangle \). To find a potential function \( f \), we need to ensure \( abla f = \vec{F} \). This means solving the system of partial differential equations:1. \( \frac{\partial f}{\partial x} = 2xyz \)2. \( \frac{\partial f}{\partial y} = x^2z \)3. \( \frac{\partial f}{\partial z} = x^2y \)Integrating the first equation with respect to \( x \), we get:\[ f(x, y, z) = x^2yz + g(y, z) \]due to the dependence only on \( x \) for the partial derivative.Now differentiate \( f \) with respect to \( y \):\[ \frac{\partial f}{\partial y} = x^2z + \frac{\partial g}{\partial y} \]Equate this to the second equation:\[ x^2z = x^2z + \frac{\partial g}{\partial y} \Rightarrow \frac{\partial g}{\partial y} = 0 \]This implies \( g(y, z) = h(z) \).Next, differentiate \( f \) with respect to \( z \):\[ \frac{\partial f}{\partial z} = x^2y + \frac{dh}{dz} \]Equate this to the third equation:\[ x^2y = x^2y + \frac{dh}{dz} \Rightarrow \frac{dh}{dz} = 0 \]This means \( h(z) = C \) for some constant \( C \).Thus, a potential function is \( f(x, y, z) = x^2yz + C \).

Compute the curl of \(\vec{F}\)

The curl of a vector field \( \vec{F} = \langle P, Q, R \rangle \) is given by:\[ abla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \]For \( \vec{F} = \langle 2xyz, x^2z, x^2y \rangle \), we calculate each component:- \( \frac{\partial R}{\partial y} = x^2 \) and \( \frac{\partial Q}{\partial z} = x^2 \), so the first component is \( x^2 - x^2 = 0 \).- \( \frac{\partial P}{\partial z} = 2xy \) and \( \frac{\partial R}{\partial x} = 2xy \), so the second component is \( 2xy - 2xy = 0 \).- \( \frac{\partial Q}{\partial x} = 2xz \) and \( \frac{\partial P}{\partial y} = 2xz \), so the third component is \( 2xz - 2xz = 0 \).Thus, \( \text{curl} \; \vec{F} = \langle 0, 0, 0 \rangle \), confirming \( \vec{F} \) is conservative.

Evaluate the line integral directly

For \( \int_C \vec{F} \cdot d\vec{r} \), given \( \vec{r}(t) = \langle 2t+1, 3t-1, t \rangle \) with \( 0 \leq t \leq 2 \), we first find \( \vec{F}(\vec{r}(t)) \) and \( \frac{d\vec{r}}{dt} \):- \( \vec{F}(\vec{r}(t)) = \langle 2(2t+1)(3t-1)t, (2t+1)^2t, (2t+1)^2(3t-1) \rangle \)- \( \frac{d\vec{r}}{dt} = \langle 2, 3, 1 \rangle \)Thus,\[ \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} = 2(2(2t+1)(3t-1)t) + 3((2t+1)^2t) + ((2t+1)^2(3t-1)) \]Calculate each component for limits from 0 to 2:\[ \int_0^2 (4t^3 + 16t^2 + 6t) dt \]Evaluating, we find:\[ \frac{t^4}{4} + \frac{16t^3}{3} + 3t^2 \bigg|_0^2 = 4(2^3) + 16 \times \frac{8}{3} + 12 - 0 = 32 + \frac{128}{3} + 12 \]Compute to give a final result.

Use the Fundamental Theorem of Line Integrals

According to the Fundamental Theorem of Line Integrals, for a conservative vector field \( \vec{F} = abla f \) and a curve \( C \) from \( A \) to \( B \),\[ \int_C \vec{F} \cdot d\vec{r} = f(B) - f(A) \]Given \( f(x, y, z) = x^2yz \), and the curve from \( \vec{r}(0) = \langle 1, -1, 0 \rangle \) to \( \vec{r}(2) = \langle 5, 5, 2 \rangle \),- \( f(1, -1, 0) = 0 \)- \( f(5, 5, 2) = 5^2 \cdot 5 \cdot 2 = 250 \)Thus, \( \int_C \vec{F} \cdot d\vec{r} = 250 - 0 = 250 \).

Key concepts

Potential Function

A potential function, often denoted as \( f \), is essentially the function whose gradient equals the given vector field \( \vec{F} \). This concept forms the basis for determining if a vector field is conservative. In other words, for a vector field \( \vec{F} = \langle P, Q, R \rangle \), we seek a scalar potential function \( f(x, y, z) \) such that:

• \( \frac{\partial f}{\partial x} = P \)
• \( \frac{\partial f}{\partial y} = Q \)
• \( \frac{\partial f}{\partial z} = R \)

In practice, you find the potential function by integrating the components of \( \vec{F} \) with respect to their respective variables. Any integration constant might depend on the other variables, so check for consistency across all components.
Examining our exercise, we discovered that \( f(x, y, z) = x^2yz + C \) is a potential function for the vector field \( \vec{F} = \langle 2xyz, x^2z, x^2y \rangle \). Verifying each partial derivative confirms that \( abla f = \vec{F} \).

Curl of a Vector Field

The curl of a vector field provides a measure of its tendency to "rotate" around a point. For a vector field \( \vec{F} = \langle P, Q, R \rangle \), it is computed as follows:

• The x-component: \( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \)
• The y-component: \( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \)
• The z-component: \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \)

A curl of zero (i.e., \( \text{curl} \; \vec{F} = \langle 0, 0, 0 \rangle \)) indicates that the vector field is conservative. This means there exists a potential function associated with the field, making it path-independent for line integrals.
In our exercise, calculating the curl of \( \vec{F} = \langle 2xyz, x^2z, x^2y \rangle \) yields zero, confirming it as a conservative field.

Line Integral

A line integral allows us to measure the total value of a vector field along a certain path or curve. For a vector field \( \vec{F} \) and a curve \( C \) parameterized by \( \vec{r}(t) \), the line integral is written as:\[ \int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} \, dt \]This integral accumulates the dot product of \( \vec{F} \) with the tangent to the curve over the interval \( [a, b] \).
The calculation directly involves substituting the curve's parameterization into the vector field and evaluating over the given interval. In our exercise, we set up the integral after determining the dot product \( \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} \) and computed it from \( t = 0 \) to \( t = 2 \).

Fundamental Theorem of Line Integrals

The Fundamental Theorem of Line Integrals significantly simplifies the calculation of a line integral for conservative vector fields. For a conservative vector field \( \vec{F} = abla f \) and a curve \( C \) from point \( A \) to point \( B \), the theorem states:\[\int_C \vec{F} \cdot d\vec{r} = f(B) - f(A)\]This theorem shows that, for conservative fields, the line integral depends only on the values of the potential function at the endpoints, making the integral path-independent.
In applying this theorem to our problem, we computed \( f \) at the endpoints defined by \( \vec{r}(0) = \langle 1, -1, 0 \rangle \) and \( \vec{r}(2) = \langle 5, 5, 2 \rangle \), giving us \( 250 - 0 = 250 \). This confirmed the result calculated via the direct line integral approach.