The line integral ∮ C f . d r directly as well as using Green’s theorem.

Question

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Answer 1

Question

Chapter 17, Problem 1CRE

To determine

To evaluate:

The line integral $\oint_{C} f . d r$ directly as well as using Green’s theorem.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The given linear integral has been evaluated to zero.

$\oint_{C} f . d r = 0$

Explanation of Solution

Given:

A multivariable vector field $f (x, y) = 〈 x + y^{2}, x^{2} - y 〉$ . A unit circle C oriented counter-clockwise.

Formula used:

$\oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t$

As per Green’s Theorem:

$\oint_{C} f . d r = \int_{C} P d x + Q d y = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A$

Calculation:

Using line integral, we can represent the circle C as:

$\begin{array}{l} C : x^{2} + y^{2} = 1 \Rightarrow x = \cos (t), y = \sin (t); 0 \leq t \leq 2 π \\ C : r (t) = 〈 \cos (t), \sin (t) 〉; 0 \leq t \leq 2 π \end{array}$

We have,

$\Rightarrow r (t) = 〈 \cos t, \sin t 〉 \Rightarrow r' (t) = 〈 - \sin t, \cos t 〉$

Thus, the line integral becomes:

$\begin{array}{l} \Rightarrow \oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} 〈 \cos t + \sin^{2} t, \cos^{2} t - \sin t 〉 \cdot 〈 - \sin t, \cos t 〉 d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin t \cos t - \sin^{3} t + \cos^{3} t - \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin^{3} t + \cos^{3} t - 2 \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = 0 \end{array}$

Now, we have to evaluate the given line integral using Green’s Theorem. Using Green’s theorem, we can write:

$\begin{array}{l} \oint_{C} f . d r = \int_{C} P d x + Q d y \\ \oint_{C} f . d r = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A \\ \oint_{C} f . d r = \iint_{D} (\frac{δ}{δ x} (x^{2} - y) - \frac{δ}{δ y} (x + y^{2})) d A \\ \oint_{C} f . d r = \iint_{D} (2 x - 2 y) d A \\ \oint_{C} f . d r = \int_{0}^{2 π} \int_{0}^{1} 2 (r \cos t - r \sin t) r d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} \int_{0}^{1} r^{2} (\cos t - \sin t) d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} (\cos t - \sin t) d t \int_{0}^{1} r^{2} d r \\ \oint_{C} f . d r = 2 (0) (\frac{1}{3}) = 0 \\ \oint_{C} f . d r = 0 \end{array}$

Conclusion:

The given integral was evaluated both using direct linear integrals as well as using Green’s Theorem.

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Answer 2

Question

Chapter 17, Problem 1CRE

To determine

To evaluate:

The line integral $\oint_{C} f . d r$ directly as well as using Green’s theorem.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The given linear integral has been evaluated to zero.

$\oint_{C} f . d r = 0$

Explanation of Solution

Given:

A multivariable vector field $f (x, y) = 〈 x + y^{2}, x^{2} - y 〉$ . A unit circle C oriented counter-clockwise.

Formula used:

$\oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t$

As per Green’s Theorem:

$\oint_{C} f . d r = \int_{C} P d x + Q d y = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A$

Calculation:

Using line integral, we can represent the circle C as:

$\begin{array}{l} C : x^{2} + y^{2} = 1 \Rightarrow x = \cos (t), y = \sin (t); 0 \leq t \leq 2 π \\ C : r (t) = 〈 \cos (t), \sin (t) 〉; 0 \leq t \leq 2 π \end{array}$

We have,

$\Rightarrow r (t) = 〈 \cos t, \sin t 〉 \Rightarrow r' (t) = 〈 - \sin t, \cos t 〉$

Thus, the line integral becomes:

$\begin{array}{l} \Rightarrow \oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} 〈 \cos t + \sin^{2} t, \cos^{2} t - \sin t 〉 \cdot 〈 - \sin t, \cos t 〉 d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin t \cos t - \sin^{3} t + \cos^{3} t - \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin^{3} t + \cos^{3} t - 2 \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = 0 \end{array}$

Now, we have to evaluate the given line integral using Green’s Theorem. Using Green’s theorem, we can write:

$\begin{array}{l} \oint_{C} f . d r = \int_{C} P d x + Q d y \\ \oint_{C} f . d r = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A \\ \oint_{C} f . d r = \iint_{D} (\frac{δ}{δ x} (x^{2} - y) - \frac{δ}{δ y} (x + y^{2})) d A \\ \oint_{C} f . d r = \iint_{D} (2 x - 2 y) d A \\ \oint_{C} f . d r = \int_{0}^{2 π} \int_{0}^{1} 2 (r \cos t - r \sin t) r d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} \int_{0}^{1} r^{2} (\cos t - \sin t) d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} (\cos t - \sin t) d t \int_{0}^{1} r^{2} d r \\ \oint_{C} f . d r = 2 (0) (\frac{1}{3}) = 0 \\ \oint_{C} f . d r = 0 \end{array}$

Conclusion:

The given integral was evaluated both using direct linear integrals as well as using Green’s Theorem.

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Answer 3

Question

Chapter 17, Problem 1CRE

To determine

To evaluate:

The line integral $\oint_{C} f . d r$ directly as well as using Green’s theorem.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The given linear integral has been evaluated to zero.

$\oint_{C} f . d r = 0$

Explanation of Solution

Given:

A multivariable vector field $f (x, y) = 〈 x + y^{2}, x^{2} - y 〉$ . A unit circle C oriented counter-clockwise.

Formula used:

$\oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t$

As per Green’s Theorem:

$\oint_{C} f . d r = \int_{C} P d x + Q d y = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A$

Calculation:

Using line integral, we can represent the circle C as:

$\begin{array}{l} C : x^{2} + y^{2} = 1 \Rightarrow x = \cos (t), y = \sin (t); 0 \leq t \leq 2 π \\ C : r (t) = 〈 \cos (t), \sin (t) 〉; 0 \leq t \leq 2 π \end{array}$

We have,

$\Rightarrow r (t) = 〈 \cos t, \sin t 〉 \Rightarrow r' (t) = 〈 - \sin t, \cos t 〉$

Thus, the line integral becomes:

$\begin{array}{l} \Rightarrow \oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} 〈 \cos t + \sin^{2} t, \cos^{2} t - \sin t 〉 \cdot 〈 - \sin t, \cos t 〉 d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin t \cos t - \sin^{3} t + \cos^{3} t - \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin^{3} t + \cos^{3} t - 2 \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = 0 \end{array}$

Now, we have to evaluate the given line integral using Green’s Theorem. Using Green’s theorem, we can write:

$\begin{array}{l} \oint_{C} f . d r = \int_{C} P d x + Q d y \\ \oint_{C} f . d r = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A \\ \oint_{C} f . d r = \iint_{D} (\frac{δ}{δ x} (x^{2} - y) - \frac{δ}{δ y} (x + y^{2})) d A \\ \oint_{C} f . d r = \iint_{D} (2 x - 2 y) d A \\ \oint_{C} f . d r = \int_{0}^{2 π} \int_{0}^{1} 2 (r \cos t - r \sin t) r d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} \int_{0}^{1} r^{2} (\cos t - \sin t) d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} (\cos t - \sin t) d t \int_{0}^{1} r^{2} d r \\ \oint_{C} f . d r = 2 (0) (\frac{1}{3}) = 0 \\ \oint_{C} f . d r = 0 \end{array}$

Conclusion:

The given integral was evaluated both using direct linear integrals as well as using Green’s Theorem.

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Answer 4

Question

Chapter 17, Problem 1CRE

To determine

To evaluate:

The line integral $\oint_{C} f . d r$ directly as well as using Green’s theorem.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The given linear integral has been evaluated to zero.

$\oint_{C} f . d r = 0$

Explanation of Solution

Given:

A multivariable vector field $f (x, y) = 〈 x + y^{2}, x^{2} - y 〉$ . A unit circle C oriented counter-clockwise.

Formula used:

$\oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t$

As per Green’s Theorem:

$\oint_{C} f . d r = \int_{C} P d x + Q d y = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A$

Calculation:

Using line integral, we can represent the circle C as:

$\begin{array}{l} C : x^{2} + y^{2} = 1 \Rightarrow x = \cos (t), y = \sin (t); 0 \leq t \leq 2 π \\ C : r (t) = 〈 \cos (t), \sin (t) 〉; 0 \leq t \leq 2 π \end{array}$

We have,

$\Rightarrow r (t) = 〈 \cos t, \sin t 〉 \Rightarrow r' (t) = 〈 - \sin t, \cos t 〉$

Thus, the line integral becomes:

$\begin{array}{l} \Rightarrow \oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} 〈 \cos t + \sin^{2} t, \cos^{2} t - \sin t 〉 \cdot 〈 - \sin t, \cos t 〉 d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin t \cos t - \sin^{3} t + \cos^{3} t - \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin^{3} t + \cos^{3} t - 2 \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = 0 \end{array}$

Now, we have to evaluate the given line integral using Green’s Theorem. Using Green’s theorem, we can write:

$\begin{array}{l} \oint_{C} f . d r = \int_{C} P d x + Q d y \\ \oint_{C} f . d r = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A \\ \oint_{C} f . d r = \iint_{D} (\frac{δ}{δ x} (x^{2} - y) - \frac{δ}{δ y} (x + y^{2})) d A \\ \oint_{C} f . d r = \iint_{D} (2 x - 2 y) d A \\ \oint_{C} f . d r = \int_{0}^{2 π} \int_{0}^{1} 2 (r \cos t - r \sin t) r d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} \int_{0}^{1} r^{2} (\cos t - \sin t) d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} (\cos t - \sin t) d t \int_{0}^{1} r^{2} d r \\ \oint_{C} f . d r = 2 (0) (\frac{1}{3}) = 0 \\ \oint_{C} f . d r = 0 \end{array}$

Conclusion:

The given integral was evaluated both using direct linear integrals as well as using Green’s Theorem.

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Answer 5

Chapter 17, Problem 1CRE

To determine

To evaluate:

The line integral $\oint_{C} f . d r$ directly as well as using Green’s theorem.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The given linear integral has been evaluated to zero.

$\oint_{C} f . d r = 0$

Explanation of Solution

Given:

A multivariable vector field $f (x, y) = 〈 x + y^{2}, x^{2} - y 〉$ . A unit circle C oriented counter-clockwise.

Formula used:

$\oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t$

As per Green’s Theorem:

$\oint_{C} f . d r = \int_{C} P d x + Q d y = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A$

Calculation:

Using line integral, we can represent the circle C as:

$\begin{array}{l} C : x^{2} + y^{2} = 1 \Rightarrow x = \cos (t), y = \sin (t); 0 \leq t \leq 2 π \\ C : r (t) = 〈 \cos (t), \sin (t) 〉; 0 \leq t \leq 2 π \end{array}$

We have,

$\Rightarrow r (t) = 〈 \cos t, \sin t 〉 \Rightarrow r' (t) = 〈 - \sin t, \cos t 〉$

Thus, the line integral becomes:

$\begin{array}{l} \Rightarrow \oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} 〈 \cos t + \sin^{2} t, \cos^{2} t - \sin t 〉 \cdot 〈 - \sin t, \cos t 〉 d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin t \cos t - \sin^{3} t + \cos^{3} t - \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin^{3} t + \cos^{3} t - 2 \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = 0 \end{array}$

Now, we have to evaluate the given line integral using Green’s Theorem. Using Green’s theorem, we can write:

$\begin{array}{l} \oint_{C} f . d r = \int_{C} P d x + Q d y \\ \oint_{C} f . d r = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A \\ \oint_{C} f . d r = \iint_{D} (\frac{δ}{δ x} (x^{2} - y) - \frac{δ}{δ y} (x + y^{2})) d A \\ \oint_{C} f . d r = \iint_{D} (2 x - 2 y) d A \\ \oint_{C} f . d r = \int_{0}^{2 π} \int_{0}^{1} 2 (r \cos t - r \sin t) r d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} \int_{0}^{1} r^{2} (\cos t - \sin t) d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} (\cos t - \sin t) d t \int_{0}^{1} r^{2} d r \\ \oint_{C} f . d r = 2 (0) (\frac{1}{3}) = 0 \\ \oint_{C} f . d r = 0 \end{array}$

Conclusion:

The given integral was evaluated both using direct linear integrals as well as using Green’s Theorem.

Answer 6

Chapter 17, Problem 1CRE

To determine

To evaluate:

The line integral $\oint_{C} f . d r$ directly as well as using Green’s theorem.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The given linear integral has been evaluated to zero.

$\oint_{C} f . d r = 0$

Explanation of Solution

Given:

A multivariable vector field $f (x, y) = 〈 x + y^{2}, x^{2} - y 〉$ . A unit circle C oriented counter-clockwise.

Formula used:

$\oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t$

As per Green’s Theorem:

$\oint_{C} f . d r = \int_{C} P d x + Q d y = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A$

Calculation:

Using line integral, we can represent the circle C as:

$\begin{array}{l} C : x^{2} + y^{2} = 1 \Rightarrow x = \cos (t), y = \sin (t); 0 \leq t \leq 2 π \\ C : r (t) = 〈 \cos (t), \sin (t) 〉; 0 \leq t \leq 2 π \end{array}$

We have,

$\Rightarrow r (t) = 〈 \cos t, \sin t 〉 \Rightarrow r' (t) = 〈 - \sin t, \cos t 〉$

Thus, the line integral becomes:

$\begin{array}{l} \Rightarrow \oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} 〈 \cos t + \sin^{2} t, \cos^{2} t - \sin t 〉 \cdot 〈 - \sin t, \cos t 〉 d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin t \cos t - \sin^{3} t + \cos^{3} t - \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin^{3} t + \cos^{3} t - 2 \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = 0 \end{array}$

Now, we have to evaluate the given line integral using Green’s Theorem. Using Green’s theorem, we can write:

$\begin{array}{l} \oint_{C} f . d r = \int_{C} P d x + Q d y \\ \oint_{C} f . d r = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A \\ \oint_{C} f . d r = \iint_{D} (\frac{δ}{δ x} (x^{2} - y) - \frac{δ}{δ y} (x + y^{2})) d A \\ \oint_{C} f . d r = \iint_{D} (2 x - 2 y) d A \\ \oint_{C} f . d r = \int_{0}^{2 π} \int_{0}^{1} 2 (r \cos t - r \sin t) r d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} \int_{0}^{1} r^{2} (\cos t - \sin t) d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} (\cos t - \sin t) d t \int_{0}^{1} r^{2} d r \\ \oint_{C} f . d r = 2 (0) (\frac{1}{3}) = 0 \\ \oint_{C} f . d r = 0 \end{array}$

Conclusion:

The given integral was evaluated both using direct linear integrals as well as using Green’s Theorem.

Answer 7

Chapter 17, Problem 1CRE

To determine

To evaluate:

The line integral $\oint_{C} f . d r$ directly as well as using Green’s theorem.

Answer 8

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

The given linear integral has been evaluated to zero.

$\oint_{C} f . d r = 0$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

A multivariable vector field $f (x, y) = 〈 x + y^{2}, x^{2} - y 〉$ . A unit circle C oriented counter-clockwise.

Formula used:

$\oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t$

As per Green’s Theorem:

$\oint_{C} f . d r = \int_{C} P d x + Q d y = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A$

Calculation:

Using line integral, we can represent the circle C as:

$\begin{array}{l} C : x^{2} + y^{2} = 1 \Rightarrow x = \cos (t), y = \sin (t); 0 \leq t \leq 2 π \\ C : r (t) = 〈 \cos (t), \sin (t) 〉; 0 \leq t \leq 2 π \end{array}$

We have,

$\Rightarrow r (t) = 〈 \cos t, \sin t 〉 \Rightarrow r' (t) = 〈 - \sin t, \cos t 〉$

Thus, the line integral becomes:

$\begin{array}{l} \Rightarrow \oint_{C} f . d r = \int_{a}^{b} f (r (t)) \cdot r' (t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} 〈 \cos t + \sin^{2} t, \cos^{2} t - \sin t 〉 \cdot 〈 - \sin t, \cos t 〉 d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin t \cos t - \sin^{3} t + \cos^{3} t - \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = \int_{0}^{2 π} (- \sin^{3} t + \cos^{3} t - 2 \sin t \cos t) d t \\ \Rightarrow \oint_{C} f . d r = 0 \end{array}$

Now, we have to evaluate the given line integral using Green’s Theorem. Using Green’s theorem, we can write:

$\begin{array}{l} \oint_{C} f . d r = \int_{C} P d x + Q d y \\ \oint_{C} f . d r = \iint_{D} (\frac{δ Q}{δ x} - \frac{δ P}{δ y}) d A \\ \oint_{C} f . d r = \iint_{D} (\frac{δ}{δ x} (x^{2} - y) - \frac{δ}{δ y} (x + y^{2})) d A \\ \oint_{C} f . d r = \iint_{D} (2 x - 2 y) d A \\ \oint_{C} f . d r = \int_{0}^{2 π} \int_{0}^{1} 2 (r \cos t - r \sin t) r d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} \int_{0}^{1} r^{2} (\cos t - \sin t) d r d t \\ \oint_{C} f . d r = 2 \int_{0}^{2 π} (\cos t - \sin t) d t \int_{0}^{1} r^{2} d r \\ \oint_{C} f . d r = 2 (0) (\frac{1}{3}) = 0 \\ \oint_{C} f . d r = 0 \end{array}$