To calculate the surface area of 3, we'll use the formule for the surface area of a parametric surface given by: Given the parameterization and then compute the cross product and its magnitude. Now, compute the cross product: Now, we set up the integral: So, the surface area of Sis 2pi^2. Explanation: To calculate the surface ares of S, we'll use the formula for the surface area of a parametric surface given by: A = = ||||ru × ru|| du dv Where r(u, v) = (x(u, v), y(u, v), z(u, v)) is the parameterization of the surface, ru and r₁, are the partial derivatives of r with respect to u and v respectively, and ||rx|| denotes the magnitude of the cross product of rμ and rv. Given the parameterization (x, y, z) = (cos z, sin z, z) for S, we need to find the partial derivatives rμ and r₁, and then compute the cross product and its magnitude. The partial derivatives are: მე მყ მ2 ru = Ju' Ju' Ju ru = მე მყ მ Əv' Iv' Əv მუ მყ მო r₁ = Iv' Iv' Iv Given (x, y, z) = (cos z, sin z, z), we have u zand v = z. So, =-sin z, = cos z, 9 = 1. Now, compute the cross product: i j k❘ ru x rv sin z cos z 0 = (-c 0 0 1 The magnitude of r xr, is: ||r, xr = √(cos z)² + (- sin z) Now, we set up the integral: A = √ždudu dv A = (√2. √ du) ( ) A=(√)(0) A = (√ √² · [— — — ( − 1 )]) · ([^ – 2 A = (√√2. π) (π) = 22 So, the surface area of S is 2π².
To calculate the surface area of 3, we'll use the formule for the surface area of a parametric surface given by: Given the parameterization and then compute the cross product and its magnitude. Now, compute the cross product: Now, we set up the integral: So, the surface area of Sis 2pi^2. Explanation: To calculate the surface ares of S, we'll use the formula for the surface area of a parametric surface given by: A = = ||||ru × ru|| du dv Where r(u, v) = (x(u, v), y(u, v), z(u, v)) is the parameterization of the surface, ru and r₁, are the partial derivatives of r with respect to u and v respectively, and ||rx|| denotes the magnitude of the cross product of rμ and rv. Given the parameterization (x, y, z) = (cos z, sin z, z) for S, we need to find the partial derivatives rμ and r₁, and then compute the cross product and its magnitude. The partial derivatives are: მე მყ მ2 ru = Ju' Ju' Ju ru = მე მყ მ Əv' Iv' Əv მუ მყ მო r₁ = Iv' Iv' Iv Given (x, y, z) = (cos z, sin z, z), we have u zand v = z. So, =-sin z, = cos z, 9 = 1. Now, compute the cross product: i j k❘ ru x rv sin z cos z 0 = (-c 0 0 1 The magnitude of r xr, is: ||r, xr = √(cos z)² + (- sin z) Now, we set up the integral: A = √ždudu dv A = (√2. √ du) ( ) A=(√)(0) A = (√ √² · [— — — ( − 1 )]) · ([^ – 2 A = (√√2. π) (π) = 22 So, the surface area of S is 2π².
Trigonometry (MindTap Course List)
10th Edition
ISBN:9781337278461
Author:Ron Larson
Publisher:Ron Larson
Chapter2: Analytic Trigonometry
Section2.3: Solving Trigonometric Equations
Problem 11ECP
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I got this answer, but part of the explanation was cut of.
could you help with the missing parts, and explain in more detail, how z=u and x= u?
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