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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Hello
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?
Transcribed Image Text: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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6314ee38-ce53-4b4c-a78d-7a20d07b3d58%2F4a45b5d6-8edf-4dee-ac30-f964140e9ce1%2F1jl4vm_processed.png&w=3840&q=75)
Transcribed Image Text: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.
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