Chapter 14: Q. 37 (page 1107)

Use the Fundamental Theorem of Line Integrals, if applicable, to evaluate the integrals in Exercises 37–44. Otherwise, show that the vector field is not conservative.
$F (x, y) = (\frac{1}{x} + \ln y, \frac{1}{y} + \frac{x}{y})$ , with C the straight line segment from $(π, e) to (1, π)$ .

Short Answer

Expert verified

The integral is $\int_{C} F (x, y) \cdot d r = \ln (π) - 1 - π$ .

Step by step solution

Step 1. Given Information

Use the Fundamental Theorem of Line Integrals, if applicable, to evaluate the integrals in the given exercises. Otherwise, show that the vector field is not conservative.

$F (x, y) = (\frac{1}{x} + \ln y, \frac{1}{y} + \frac{x}{y})$ , with C the straight line segment from $(π, e) to (1, π)$ .

Step 2. Firstly checking the given field is conservative or not.

$\frac{d F (x, y)}{d y} = \frac{d}{d y} (\frac{1}{x} + \ln y) \frac{d F (x, y)}{d x} = \frac{d}{d x} (\frac{1}{y} + \frac{x}{y}) \frac{d F (x, y)}{d y} = \frac{d}{d y} \frac{1}{x} + \frac{d}{d y} \ln y \frac{d F (x, y)}{d x} = \frac{d}{d x} \frac{1}{y} + \frac{d}{d x} \frac{x}{y} \frac{d F (x, y)}{d y} = 0 + \frac{1}{y} \frac{d F (x, y)}{d x} = 0 + \frac{1}{y} \frac{d F (x, y)}{d y} = \frac{1}{y} \frac{d F (x, y)}{d x} = \frac{1}{y}$

Since, $\frac{d F (x, y)}{d y} = \frac{d F (x, y)}{d x}$ , so the given function is conservative.

Step 3. The given function is F(x,y)=1x+lny,1y+xy

The points are $(π, e) to (1, π)$

We find a potential function for F:

$f (x, y) = \int (\frac{1}{y} + \frac{x}{y}) d y f (x, y) = \int \frac{1}{y} d y + x \int \frac{1}{y} d y f (x, y) = \ln (y) + x \ln (y)$

By the Fundamental Theorem,

$\int_{C} F (x, y) \cdot d r = f (1, π) - f (π, e)$

Step 4. Now solving ∫CF(x,y)·dr=f(1,π)−f(π,e)

$\int_{C} F (x, y) \cdot d r = [\ln (π) + 1 \ln (π)] - [\ln (e) + πln (e)] \int_{C} F (x, y) \cdot d r = \ln (π) + \ln (π) - [1 + π \cdot 1] \int_{C} F (x, y) \cdot d r = 2 \ln (π) - [1 + π] \int_{C} F (x, y) \cdot d r = 2 \ln (π) - 1 - π \int_{C} F (x, y) \cdot d r = \ln (π) - 1 - π$

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Use the Fundamental Theorem of Line Integrals, if applicable, to evaluate the integrals in Exercises 37–44. Otherwise, show that the vector field is not conservative.
$F (x, y) = (\frac{1}{x} + \ln y, \frac{1}{y} + \frac{x}{y})$ , with C the straight line segment from $(π, e) to (1, π)$ .

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Firstly checking the given field is conservative or not.

Step 3. The given function is F(x,y)=1x+lny,1y+xy

Step 4. Now solving ∫CF(x,y)·dr=f(1,π)−f(π,e)

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Use the Fundamental Theorem of Line Integrals, if applicable, to evaluate the integrals in Exercises 37–44. Otherwise, show that the vector field is not conservative.F(x,y)=1x+lny,1y+xy, with C the straight line segment from (π,e)to(1,π).

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Firstly checking the given field is conservative or not.

Step 3. The given function is F(x,y)=1x+lny,1y+xy

Step 4. Now solving ∫CF(x,y)·dr=f(1,π)−f(π,e)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Calculus

Probability and Statistics

Discrete Mathematics

Decision Maths

Statistics

Study anywhere. Anytime. Across all devices.

Use the Fundamental Theorem of Line Integrals, if applicable, to evaluate the integrals in Exercises 37–44. Otherwise, show that the vector field is not conservative.
$F (x, y) = (\frac{1}{x} + \ln y, \frac{1}{y} + \frac{x}{y})$ , with C the straight line segment from $(π, e) to (1, π)$ .