Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Find the work done by the vector field
$F (x, y) = (\cos (x^{2}) + 4 x y^{2}) i + (2^{y} - 4 x^{2} y) j$
in moving an object around the unit circle, starting and ending at $(1, 0)$ .

Ans: The required work done is $- 2 π$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information:

$F (x, y) = (\cos (x^{2}) + 4 x y^{2}) i + (2^{y} - 4 x^{2} y) j$

Step 2. Finding the work done by the vector filed:

The work done by a vector field $F$ moving a particle along the curve C is computed by the following line integral:

$\int_{C} F \cdot d r .$

Hence, evaluate the line integral $\int_{C} F \cdot d r$ to evaluate the required work done.

Step 3. Using Green's Theorem to evaluate the required the line integral:

Green's Theorem states that,

"Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by $r (t) = ⟨ (x (t), y (t)) ⟩$ for $a \leq t \leq b$ .

If a vector field $F (x, y) = <F_{1} (x, y), F_{2} (x, y)>$ is defined on R, then,

\int_{C} F \cdot d r = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A . " \dots . . (1)

Step 4. Finding the vector:

For the vector field $F (x, y) = (\cos (x^{2}) + 4 x y^{2}) i + (2^{y} - 4 x^{2} y) j$ , $F_{1} (x, y) = \cos (x^{2}) + 4 x y^{2}$ and $F_{2} (x, y) = 2^{y} - 4 x^{2} y .$

Now, first, find $\frac{\partial F_{2}}{\partial x}$ and $\frac{\partial F_{1}}{\partial y}$ .

Then,

$\frac{\partial F_{2}}{\partial x} = \frac{\partial}{\partial x} (2^{y} - 4 x^{2} y) = - 8 x y$

and,

$\frac{\partial F_{1}}{\partial y} = \frac{\partial}{\partial y} (\cos (x^{2}) + 4 x y^{2}) = 8 x y$

Step 5. Using Green's Theorem (1) to evaluate the integral:

Now, use Green's Theorem (1) to evaluate the integral $\int_{C} F \cdot d r$ as follows:

$\int_{C} F \cdot d r = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A = \iint_{R} (- 8 x y - 8 x y) d A = \iint_{R} (- 16 x y) d A \dots \dots (2)$

Step 6. Changing this integral (2) to polar coordinates:

Here, the boundary curve C is a unit circle, starting and ending at $(1, 0)$ . In polar coordinates, the region of integration is described as follows,

$R = {(r, θ) ∣ 0 \leq r \leq 1, 0 \leq θ \leq 2 π} .$

In this case,

$x = r \cos θ, y = r \sin θ, x^{2} + y^{2} = r^{2}, and d A = r d r d θ$

Step 7. Evaluating the integral (2):

Then, evaluate the integral (2) as follows:

$\int_{C} F \cdot d r = \iint_{R} (- 16 x y) d A = \int_{0}^{2 π} \int_{0}^{1} - 16 (r \cos θ) (r \sin θ) r d r d θ = \int_{0}^{2 π} \int_{0}^{1} - 16 r^{3} \sin θ \cos θ d r d θ = \int_{0}^{2 π} \sin θ \cos θ [\int_{0}^{1} - 16 r^{3} d r] d θ = \int_{0}^{2 π} \sin θ \cos θ {[- 4 r^{4}]}_{0}^{1} d θ = \int_{0}^{2 π} \sin θ \cos θ [(- 4 \cdot 1^{4}) - (- 4 \cdot 0^{4})] d θ = \int_{0}^{2 π} - 4 \sin θ \cos θ d θ = {[2 \cos^{2} θ]}_{0}^{2 π} = 2 \cos^{2} 2 π - 2 \cos^{2} 0 = 2 (1)^{2} - 2 (1)^{2} = 0 .$
Therefore, the required work done is 0.

Step 8. Correcting the Problem of the textbook:

In Book, this problem is given WRONG.

Correct Problem:

Consider the following vector field:

F (x, y) = (\cos (x^{2}) + 4 x^{2} y) i + (2^{y} - 4 x y^{2}) j

Step 9. Rewriting the Step (4) with the correct problem by finding the vector:

For the vector field $F (x, y) = (\cos (x^{2}) + 4 x^{2} y) i + (2^{y} - 4 x y^{2}) j$

$F_{1} (x, y) = \cos (x^{2}) + 4 x^{2} y$ and $F_{2} (x, y) = 2^{y} - 4 x y^{2}$

Now, first find $\frac{\partial F_{2}}{\partial x}$ and $\frac{\partial F_{1}}{\partial y}$ .

Then,

$\frac{\partial F_{2}}{\partial x} = \frac{\partial}{\partial x} (2^{y} - 4 x y^{2}) = - 4 y^{2}$

and,

$\frac{\partial F_{1}}{\partial y} = \frac{\partial}{\partial y} (\cos (x^{2}) + 4 x^{2} y) = 4 x^{2}$

Step 10. Rewriting the Step (5) with the correct problem by Using Green's Theorem (1) to evaluate the integral:

Now, use Green's Theorem (1) to evaluate the integral $\int_{C} F \cdot d r$ as follows:

$\int_{C} F \cdot d r = \iint_{R} (\frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}) d A = \iint_{R} (- 4 y^{2} - 4 x^{2}) d A = \iint_{R} - 4 (x^{2} + y^{2}) d A \dots \dots (3)$

Step 11. Rewriting the Step (6) with the correct problem by Changing this integral (3) to polar coordinates:

Change this integral (3) to polar coordinates, and integrate it.

Here, the boundary curve C is a unit circle, starting and ending at $(1, 0)$ . In polar coordinates, the region of integration is described as follows,

$R = {(r, θ) ∣ 0 \leq r \leq 1, 0 \leq θ \leq 2 π} .$

In this case,

$x = r \cos θ, y = r \sin θ, x^{2} + y^{2} = r^{2}, and d A = r d r d θ$

Step 12. Rewriting the Step (7) with the correct problem by evaluating the integral (3):

Then, evaluate the integral (3) as follows:

$\int_{C} F \cdot d r = \iint_{R} - 4 (x^{2} + y^{2}) d A = \int_{0}^{2 π} \int_{0}^{1} - 4 r^{2} \cdot r d r d θ = \int_{0}^{2 π} \int_{0}^{1} - 4 r^{3} d r d θ = \int_{0}^{2 π} [\int_{0}^{1} - 4 r^{3} d r] d θ = \int_{0}^{2 π} {[- r^{4}]}_{0}^{1} d θ = \int_{0}^{2 π} [(- 1^{4}) - (- 0^{4})] d θ = \int_{0}^{2 π} - 1 d θ = [- θ]_{0}^{2 π} = (- 2 π) - (- 0) = - 2 π .$

Therefore, the required work done is $- 2 π$ .

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Calculus

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Short Answer

Find the work done by the vector field
$F (x, y) = (\cos (x^{2}) + 4 x y^{2}) i + (2^{y} - 4 x^{2} y) j$
in moving an object around the unit circle, starting and ending at $(1, 0)$ .

Step by Step Solution

Step 1. Given information:

Step 2. Finding the work done by the vector filed:

Step 3. Using Green's Theorem to evaluate the required the line integral:

Step 4. Finding the vector:

Step 5. Using Green's Theorem (1) to evaluate the integral:

Step 6. Changing this integral (2) to polar coordinates:

Step 7. Evaluating the integral (2):

Step 8. Correcting the Problem of the textbook:

Step 9. Rewriting the Step (4) with the correct problem by finding the vector:

Step 10. Rewriting the Step (5) with the correct problem by Using Green's Theorem (1) to evaluate the integral:

Step 11. Rewriting the Step (6) with the correct problem by Changing this integral (3) to polar coordinates:

Step 12. Rewriting the Step (7) with the correct problem by evaluating the integral (3):

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Calculus

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Short Answer

Find the work done by the vector fieldF(x,y)=cosx2+4xy2i+2y-4x2yjin moving an object around the unit circle, starting and ending at (1,0).

Step by Step Solution

Step 1. Given information:

Step 2. Finding the work done by the vector filed:

Step 3. Using Green's Theorem to evaluate the required the line integral:

Step 4. Finding the vector:

Step 5. Using Green's Theorem (1) to evaluate the integral:

Step 6. Changing this integral (2) to polar coordinates:

Step 7. Evaluating the integral (2):

Step 8. Correcting the Problem of the textbook:

Step 9. Rewriting the Step (4) with the correct problem by finding the vector:

Step 10. Rewriting the Step (5) with the correct problem by Using Green's Theorem (1) to evaluate the integral:

Step 11. Rewriting the Step (6) with the correct problem by Changing this integral (3) to polar coordinates:

Step 12. Rewriting the Step (7) with the correct problem by evaluating the integral (3):

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Find the work done by the vector field
$F (x, y) = (\cos (x^{2}) + 4 x y^{2}) i + (2^{y} - 4 x^{2} y) j$
in moving an object around the unit circle, starting and ending at $(1, 0)$ .