6.3 Conservative vector fields (Page 8/16)

Calculus volume 3 Page 8 / 16

Determine whether $F (x, y) = ⟨ sin x cos y, cos x sin y ⟩$ is conservative.

It is conservative.

When using [link] , it is important to remember that a theorem is a tool, and like any tool, it can be applied only under the right conditions. In the case of [link] , the theorem can be applied only if the domain of the vector field is simply connected.

To see what can go wrong when misapplying the theorem, consider the vector field from [link] :

F (x, y) = \frac{y}{x^{2} + y^{2}} i + \frac{− x}{x^{2} + y^{2}} j .

This vector field satisfies the cross-partial property, since

\frac{\partial}{\partial y} (\frac{y}{x^{2} + y^{2}}) = \frac{(x^{2} + y^{2}) - y (2 y)}{{(x^{2} + y^{2})}^{2}} = \frac{x^{2} - y^{2}}{{(x^{2} + y^{2})}^{2}}

and

\frac{\partial}{\partial x} (\frac{− x}{x^{2} + y^{2}}) = \frac{− (x^{2} + y^{2}) + x (2 x)}{{(x^{2} + y^{2})}^{2}} = \frac{x^{2} - y^{2}}{{(x^{2} + y^{2})}^{2}} .

Since F satisfies the cross-partial property, we might be tempted to conclude that F is conservative. However, F is not conservative. To see this, let

r (t) = ⟨ cos t, sin t ⟩, 0 \leq t \leq π

be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this $C_{1})$ and let

s (t) = ⟨ cos t, − sin t ⟩, 0 \leq t \leq π

be a parameterization of the lower half of a unit circle oriented clockwise (denote this $C_{2}) .$ Notice that $C_{1}$ and $C_{2}$ have the same starting point and endpoint. Since ${sin}^{2} t + {cos}^{2} t = 1,$

F (r (t)) • r^{'} (t) = ⟨ sin (t), − cos (t) ⟩ • ⟨ − sin (t), cos (t) ⟩ = −1

and

\begin{matrix} F (s (t)) \cdot s' (t) & = ⟨ − sin t, − cos t ⟩ \cdot ⟨ − sin t, − cos t ⟩ \\ = {sin}^{2} t + {cos}^{2} t \\ = 1. \end{matrix}

Therefore,

\int_{C_{1}} F \cdot d r = \int_{0}^{π} −1 d t = − π and \int_{C_{2}} F \cdot d r = \int_{0}^{π} 1 d t = π .

Thus, $C_{1}$ and $C_{2}$ have the same starting point and endpoint, but $\int_{C_{1}} F \cdot d r \neq \int_{C_{2}} F \cdot d r .$ Therefore, F is not independent of path and F is not conservative.

To summarize: F satisfies the cross-partial property and yet F is not conservative. What went wrong? Does this contradict [link] ? The issue is that the domain of F is all of $ℝ^{2}$ except for the origin. In other words, the domain of F has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected, [link] does not apply to F .

We close this section by looking at an example of the usefulness of the Fundamental Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always decide whether the Fundamental Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form $\int_{C} F \cdot d r,$ then our first question should be: Is F conservative? If the answer is yes, then we should find a potential function and use the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, then the Fundamental Theorem for Line Integrals can’t help us and we have to use other methods, such as using [link] .

Using the fundamental theorem for line integrals

Calculate line integral $\int_{C} F \cdot d r,$ where $F (x, y, z) = ⟨ 2 x e^{y} z + e^{x} z, x^{2} e^{y} z, x^{2} e^{y} + e^{x} ⟩$ and C is any smooth curve that goes from the origin to $(1, 1, 1) .$

Before trying to compute the integral, we need to determine whether F is conservative and whether the domain of F is simply connected. The domain of F is all of $ℝ^{3},$ which is connected and has no holes. Therefore, the domain of F is simply connected. Let

P (x, y, z) = 2 x e^{y} z + e^{x} z, Q (x, y, z) = x^{2} e^{y} z, and R (x, y, z) = x^{2} e^{y} + e^{x}

so that $F = ⟨ P, Q, R ⟩ .$ Since the domain of F is simply connected, we can check the cross partials to determine whether F is conservative. Note that

\begin{matrix} P_{y} & = & 2 x e^{y} z = Q_{x} \\ P_{z} & = & 2 x e^{y} + e^{x} = R_{x} \\ Q_{z} & = & x^{2} e^{y} = R_{y} . \end{matrix}

Therefore, F is conservative.

To evaluate $\int_{C} F \cdot d r$ using the Fundamental Theorem for Line Integrals, we need to find a potential function $f$ for F . Let $f$ be a potential function for F . Then, $∇ f = F,$ and therefore $f_{x} = 2 x e^{y} z + e^{x} z .$ Integrating this equation with respect to x gives $f (x, y, z) = x^{2} e^{y} z + e^{x} z + h (y, z)$ for some function h . Differentiating this equation with respect to y gives $x^{2} e^{y} z + h_{y} = Q = x^{2} e^{y} z,$ which implies that $h_{y} = 0 .$ Therefore, h is a function of z only, and $f (x, y, z) = x^{2} e^{y} z + e^{x} z + h (z) .$ To find h , note that $f_{z} = x^{2} e^{y} + e^{x} + h' (z) = R = x^{2} e^{y} + e^{x} .$ Therefore, $h' (z) = 0$ and we can take $h (z) = 0 .$ A potential function for F is $f (x, y, z) = x^{2} e^{y} z + e^{x} z .$

Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,

\begin{matrix} \int_{C} F \cdot d r & = \int_{C} ∇ f \cdot d r \\ = f (1, 1, 1) - f (0, 0, 0) \\ = 2 e . \end{matrix}

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Practice Key Terms 6

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask

	12 AP Key Terms 12 The Nervous System By OpenStax Start Key Terms
	4 Physiotherapy Flashcards Set 4 By Rhodes Start Flashcards
	1 Microeconomics 01 What Is Economics? By OpenStax Start Flashcards
©flickr: Steve	C Programming Language By JavaChamp Team Start Quiz
	2 Neuroscience Exam 2004 3 By David Corey Start Exam
	17 AP Key Terms 17 Endocrine System By OpenStax Start Key Terms
	Clinical Psychology MCQ By Saylor Foundation Start Quiz
	Art Reviewer MCQ By Edgar Delgado Start Quiz
	19 Biology 19 The Evolution of Populations MCQ By OpenStax Start Quiz
	2 Muscular System MCQ By Nick Swain Start Quiz