The orientation of the ∂ S , that is top and bottom circles. The diagram: Applying the right hand rule , since we already have the normal in the outward direction so the compatible orientation of the top and bottom parts of the boundary of the surface eventually come out to be opposite. The top one has a counter clockwise orientation while the bottom ha clockwise orientation when viewed from above. The diagram showing the orientation is:

Question

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Answer 1

Question

Chapter 18.2, Problem 17E

(a)

To determine

To sketch:

The orientation of the $\partial S$ , that is top and bottom circles.

The diagram:

Applying the right hand rule , since we already have the normal in the outward direction so the compatible orientation of the top and bottom parts of the boundary of the surface eventually come out to be opposite. The top one has a counter clockwise orientation while the bottom ha clockwise orientation when viewed from above.

The diagram showing the orientation is:

(b)

To determine

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9

Expert Solution & Answer

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See solution

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Answer 2

Question

Chapter 18.2, Problem 17E

(a)

To determine

To sketch:

The orientation of the $\partial S$ , that is top and bottom circles.

The diagram:

Applying the right hand rule , since we already have the normal in the outward direction so the compatible orientation of the top and bottom parts of the boundary of the surface eventually come out to be opposite. The top one has a counter clockwise orientation while the bottom ha clockwise orientation when viewed from above.

The diagram showing the orientation is:

(b)

To determine

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 18.2, Problem 17E

(a)

To determine

To sketch:

The orientation of the $\partial S$ , that is top and bottom circles.

The diagram:

Applying the right hand rule , since we already have the normal in the outward direction so the compatible orientation of the top and bottom parts of the boundary of the surface eventually come out to be opposite. The top one has a counter clockwise orientation while the bottom ha clockwise orientation when viewed from above.

The diagram showing the orientation is:

(b)

To determine

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 18.2, Problem 17E

(a)

To determine

To sketch:

The orientation of the $\partial S$ , that is top and bottom circles.

The diagram:

Applying the right hand rule , since we already have the normal in the outward direction so the compatible orientation of the top and bottom parts of the boundary of the surface eventually come out to be opposite. The top one has a counter clockwise orientation while the bottom ha clockwise orientation when viewed from above.

The diagram showing the orientation is:

(b)

To determine

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 18.2, Problem 17E

(a)

To determine

To sketch:

The orientation of the $\partial S$ , that is top and bottom circles.

The diagram:

Applying the right hand rule , since we already have the normal in the outward direction so the compatible orientation of the top and bottom parts of the boundary of the surface eventually come out to be opposite. The top one has a counter clockwise orientation while the bottom ha clockwise orientation when viewed from above.

The diagram showing the orientation is:

(b)

To determine

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9

Answer 6

Chapter 18.2, Problem 17E

(a)

To determine

To sketch:

The orientation of the $\partial S$ , that is top and bottom circles.

The diagram:

Applying the right hand rule , since we already have the normal in the outward direction so the compatible orientation of the top and bottom parts of the boundary of the surface eventually come out to be opposite. The top one has a counter clockwise orientation while the bottom ha clockwise orientation when viewed from above.

The diagram showing the orientation is:

Answer 7

Chapter 18.2, Problem 17E

(a)

To determine

To sketch:

The orientation of the $\partial S$ , that is top and bottom circles.

The diagram:

Applying the right hand rule , since we already have the normal in the outward direction so the compatible orientation of the top and bottom parts of the boundary of the surface eventually come out to be opposite. The top one has a counter clockwise orientation while the bottom ha clockwise orientation when viewed from above.

The diagram showing the orientation is:

Answer 8

(b)

To determine

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9

Answer 9

(b)

To determine

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

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See solution

Answer 13

Ch. 18.1 - Prob. 1PQ Ch. 18.1 - Prob. 2PQ Ch. 18.1 - Prob. 3PQ Ch. 18.1 - Prob. 4PQ Ch. 18.1 - Prob. 5PQ Ch. 18.1 - Prob. 1E Ch. 18.1 - Prob. 2E Ch. 18.1 - Prob. 3E Ch. 18.1 - Prob. 4E Ch. 18.1 - Prob. 5E

Videos

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9

Want to see the full answer?

Chapter 18 Solutions

Videos

To verify: Stokes’ Theorem for S and F= <yz2,0,0> ∮CF→·dR→= −9

Want to see the full answer?

Chapter 18 Solutions

To verify:

Stokes’ Theorem for S and F= <yz²,0,0>

$\oint_{C} \vec{F} \cdot d \vec{R}$ = −9