Explanation Calculation: Consider the given the torus can be expressed parametrically as x = a + b cos v cos u , y = a + b cos v sin u and z = b sin v . The given parametric equations can be expressed in vector form. r u , v = x u , v i + y u , v j + z u , v k r = a + b cos v cos u i + a + b cos v sin u j + b sin v k Now, find the partial derivative of r with respect to u . ∂ r ∂ u = ∂ ∂ u a + b cos v cos u i + a + b cos v sin u j + b sin v k = ∂ ∂ u a + b cos v cos u i + ∂ ∂ u a + b cos v sin u j + ∂ ∂ u b sin v k = − a + b cos v sin u i + a + b cos v cos u j + 0 k = − a + b cos v sin u i + a + b cos v cos u j Find the partial derivative of r with respect to v . ∂ r ∂ v = ∂ ∂ v a + b cos v cos u i + a + b cos v sin u j + b sin v k = ∂ ∂ v a + b cos v cos u i + ∂ ∂ v a + b cos v sin u j + ∂ ∂ v b sin v k = 0 − b sin v cos u i + 0 − b sin v sin u j + b cos v k = − b sin v cos u i − b sin v sin u j + b cos v k Now, find the cross product of ∂ r ∂ u and ∂ r ∂ v . ∂ r ∂ u × ∂ r ∂ v = i j k − a + b cos v sin u a + b cos v cos u 0 − b sin v cos u − b sin v sin u b cos v = i a + b cos v b cos v cos u − 0 − j − a + b cos v b sin u cos v − 0 + k b a + b cos v sin 2 u sin v + b a + b cos v cos 2 u sin v = a + b cos v b cos v cos u i + a + b cos v b sin u cos v j + b a + b cos v sin v sin 2 u + cos 2 u k Use the trigonometric identity sin 2 θ + cos 2 θ = 1 . ∂ r ∂ u × ∂ r ∂ v = a + b cos v b cos v cos u i + a + b cos v b sin u cos v j + b a + b cos v sin v 1 k = a + b cos v b cos v cos u i + a + b cos v b sin u cos v j + b a + b cos v sin v k Now, find the magnitude of ∂ r ∂ u × ∂ r ∂ v

11th Edition

ISBN: 9781119244912

Author: Anton

Publisher: WILEY CONS

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Chapter 14.4, Problem 55ES

To determine

To calculate: The surface area of the torus, the torus can be expressed parametrically as x=a+bcosvcosu , y=a+bcosvsinu , and z=bsinv where u and v are the parameters 0≤u≤2π , 0≤v≤2π .

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Chapter 14.4, Problem 54ES

Chapter 14.4, Problem 56ES

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Chapter 14 Solutions

EBK CALCULUS:EARLY TRANSCENDENTALS

Ch. 14.1 - Prob. 1QCE Ch. 14.1 - The iterated integral 1524fx,ydxdy integrates f...Ch. 14.1 - Supply the missing integrand and limits of...Ch. 14.1 - Prob. 4QCE Ch. 14.1 - Evaluate the iterated integrals. 0102x+3dydx Ch. 14.1 - Evaluate the iterated integrals. 13112x4ydydx Ch. 14.1 - Evaluate the iterated integrals. 2401x2ydxdy Ch. 14.1 - Evaluate the iterated integrals. 2012x2+y2dxdy Ch. 14.1 - Evaluate the iterated integrals. 0ln30ln2ex+ydydx Ch. 14.1 - Evaluate the iterated integrals. 0201ysinxdydx

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The surface area of the torus, the torus can be expressed parametrically as x = a + b cos v cos u , y = a + b cos v sin u , and z = b sin v where u and v are the parameters 0 ≤ u ≤ 2 π , 0 ≤ v ≤ 2 π .

To calculate: The surface area of the torus, the torus can be expressed parametrically as x=a+bcosvcosu , y=a+bcosvsinu , and z=bsinv where u and v are the parameters 0≤u≤2π , 0≤v≤2π .

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Chapter 14 Solutions