How to parameterize these surfaces?

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Homework Help Overview

The discussion revolves around calculating a surface integral over the upper part of a hemisphere with radius \( a \), specifically focusing on the parameterization of the surface and the expression for \( ds \). The problem is situated within the context of multivariable calculus and surface integrals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various parameterizations for the hemisphere, questioning their correctness and the implications of different choices. There is a focus on understanding the expression for \( ds \) and the role of the normal vector \( N \) in the context of the surface integral.

Discussion Status

Some participants express confidence in the proposed parameterizations, while others seek clarification on how to derive \( ds \) and the significance of \( N \). There is an ongoing exploration of the necessary components for setting up the integral, with no clear consensus yet on the best approach.

Contextual Notes

Participants note potential confusion with similar threads and the necessity of distinguishing between different parameterizations. The discussion highlights the importance of correctly expressing \( ds \) in relation to the chosen parameters.

Sho Kano
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Homework Statement


Calculate ##\iint { y+{ z }^{ 2 }ds } ## where the surface is the upper part of a hemisphere with radius a centered at the origin with ##x\ge 0##

Homework Equations


Parameterizations:
##\sigma =\left< asin\phi cos\theta ,asin\phi sin\theta ,acos\phi \right> ,0\le \phi \le \frac { \pi }{ 2 } ,\frac { -\pi }{ 2 } \le \theta \le \frac { \pi }{ 2 } \\ N=(asin\phi )\sigma \\ \left| N \right| ={ a }^{ 2 }sin\phi \\ \\ \alpha =\left< rcos\theta ,rsin\theta ,0 \right> ,0\le r\le a,\frac { -\pi }{ 2 } \le \theta \le \frac { \pi }{ 2 } \\ N=-k\\ \left| N \right| =1##

The Attempt at a Solution


are these the right parameterizations?
 
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Check for yourself: ##\ \sigma =\left< a\sin\phi \cos\theta ,a\sin\phi \sin\theta ,a\cos\phi \right> \ ##seems right to me. For ##\ \iint ds\ ## you would then get ##\ \pi a^2, \ ## right ?

It is not clear to me what you do to express ##\ ds \ ##. What is ##N## and what is the function of ##N## ?
 
BvU said:
Check for yourself: ##\ \sigma =\left< a\sin\phi \cos\theta ,a\sin\phi \sin\theta ,a\cos\phi \right> \ ##seems right to me. For ##\ \iint ds\ ## you would then get ##\ \pi a^2, \ ## right ?

It is not clear to me what you do to express ##\ ds \ ##. What is ##N## and what is the function of ##N## ?
Oh sorry, by the integral I mean a surface integral. N is the normal. Both parameterizations seem right to me...i originally had ##a## instead of ##r## for the second parameterization. But that would just give me a circle, not a disk (a surface)
 
Are we mixing up two threads with almost the same title ?
Not clear to me why you need ##N## in this thread. But you sure need ##ds## and I haven't seen how you are going to express that in the parameters
 

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