Explanation Given: Surface S represented by the parametric curve G ( u , v ) = ( ( 1 + v cos u 2 ) cos u , ( 1 + v cos u 2 ) sin u , v sin u 2 ) ; 0 ≤ u ≤ 2 π , − 1 2 ≤ v ≤ 1 2 N ( u , 0 ) = ( cos u sin u 2 , sin u sin u 2 , − cos u 2 ) is a unit vector that varies continuously but that N ( 2 π , 0 ) = − N ( 0 , 0 ) shows that the surface S is not orientable. Proof: Here the Surface represented by the parametric curve G ( u , v ) = ( ( 1 + v cos u 2 ) cos u , ( 1 + v cos u 2 ) sin u , v sin u 2 ) ; 0 ≤ u ≤ 2 π , − 1 2 ≤ v ≤ 1 2 in the form of frustum of cone and thereby the lower part of the surface would be circle, i.e., cross sectional part of the bottom of the solid would be a circle. In the parametric curve G ( u , v ) = ( ( 1 + v cos u 2 ) cos u , ( 1 + v cos u 2 ) sin u , v sin u 2 ) ; 0 ≤ u ≤ 2 π , − 1 2 ≤ v ≤ 1 2 The x , y and z components are x = ( 1 + v cos u 2 ) cos u = cos u + v cos u cos u 2 y = ( 1 + v cos u 2 ) sin u = sin u + v cos u 2 sin u z = v sin u 2 where, 0 ≤ u ≤ 2 π , − 1 2 ≤ v ≤ 1 2 Now for xy -plane z = 0 ⇒ v sin u 2 = 0 ⇒ v = 0 or sin u 2 = 0 ⇒ v = 0 or u 2 = n π ⇒ v = 0 or u = 2 n π Now, u = 2 n π ⇒ u = 2 π , 4 π , .. , 2 n π Since, 0 ≤ u ≤ 2 π , − 1 2 ≤ v ≤ 1 2 ⇒ u ≠ 2 n π ⇒ v = 0 Now, for v = 0, the equation of surface becomes G ( u , 0 ) = ( ( 1 + 0 × cos u 2 ) cos u , ( 1 + 0 × cos u 2 ) sin u , 0 × sin u 2 ) ⇒ G ( u , 0 ) = ( cos u , sin u , 0 ) Also, ‖ G ( u , 0 ) ‖ = ‖ ( cos u , sin u , 0 ) ‖ = cos 2 u + sin 2 u + 0 = 1 Thus, the intersection of the surface S with the xy -plane is the unit circle given by G ( u , 0 ) = ( cos u , sin u , 0 ) . Now, we need to determine the vector normal to the surface which is given by n → = ∂ ∂ u ( G ( u , v ) ) × ∂ ∂ v ( G ( u , v ) ) ⇒ n → = ∂ ∂ u ( ( ( 1 + v cos u 2 ) cos u , ( 1 + v cos u 2 ) sin u , v sin u 2 ) ) × ∂ ∂ v ( ( ( 1 + v cos u 2 ) cos u , ( 1 + v cos u 2 ) sin u , v sin u 2 ) ) ⇒ n → = ∂ ∂ u ( ( cos u + v cos u cos u 2 , sin u + v sin u cos u 2 , v sin u 2 ) ) × ∂ ∂ v ( ( cos u + v cos u cos u 2 , sin u + v sin u cos u 2 , v sin u 2 ) ) ⇒ n → = ( − sin u + v ( − cos u 2 sin u − 1 2 cos u sin u 2 ) , cos u + v ( cos u 2 cos u − 1 2 sin u sin u 2 ) , v 2 cos u 2 ) × ( cos u 2 cos u , cos u 2 sin u , sin u 2 ) ⇒ n → = ( − sin u + ( − v cos u 2 sin u − v 2 cos u sin u 2 ) , cos u + ( v cos u 2 cos u − v 2 sin u sin u 2 ) , v 2 cos u 2 ) × ( cos u 2 cos u , cos u 2 sin u , sin u 2 ) ⇒ n → = ( { ( cos u + ( v cos u 2 cos u − v 2 sin u sin u 2 ) ) sin u 2 − ( v 2 cos u 2 ) ( cos u 2 sin u ) } , { ( v 2 cos u 2 ) cos u 2 cos u − ( − sin u + ( − v cos u 2 sin u − v 2 cos u sin u 2 ) ) sin u 2 } , { ( − sin u + ( − v cos u 2 sin u − v 2 cos u sin u 2 ) ) cos u 2 sin u − ( cos u + ( v cos u 2 cos u − v 2 sin u sin u To determine (b) To Calculate: The surface area of given surface S . To determine (c) To Calculate: The given surface integral.

4th Edition

ISBN: 9781319055844

Author: FRANZOSA

Publisher: VST

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Question

Chapter 17.5, Problem 38E

To determine

(a)

To Verify:

The given norm condition.

To determine

(b)

To Calculate:

The surface area of given surface S.

To determine

(c)

To Calculate:

The given surface integral.

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Chapter 17.5, Problem 37E

Chapter 17, Problem 1CRE

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Chapter 17 Solutions

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(a) To Verify: The given norm condition.

Solution Summary: The author shows that the surface S is not orientable. The parametric curve G is a unit vector that varies continuously.

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(a) To Verify: The given norm condition.

(b) To Calculate: The surface area of given surface S.

(c) To Calculate: The given surface integral.

Want to see the full answer?

Chapter 17 Solutions

(a)

To Verify:

The given norm condition.

(b)

To Calculate:

The surface area of given surface S.

(c)

To Calculate:

The given surface integral.