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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: Marina.nascx
Type : Application
Page(s) : 6
Taille Size: 39.80 Ko KB
Mis en ligne Uploaded: 28/05/2017 - 21:20:58
Uploadeur Uploader: Marina.nascx (Profil)
Téléchargements Downloads: 8
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a991314

Description 

252 CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15

13. 1 represents the solid sphere of radius 1 centered at the origin.
3 3
4
restricts the solid to that portion on or below the cone = 4
.




2 2 2
15. 2 + 2 because the solid lies above the cone. Squaring both sides of this inequality gives +
2 2 2 2 2 2 2
2 + + = = cos2 1 2
2
cos2 1
2
. The cone opens upward so that the inequality is
1 2 2 2 2
cos 2
, or equivalently 0 4
. In spherical coordinates the sphere = + + is cos =

= cos . 0 cos because the solid lies below the sphere. The solid can therefore be described as the region in
spherical coordinates satisfying 0 cos , 0 4
.

17. The region of integration is given in spherical coordinates by
= {( )|0 3 0 2 0 6}. This represents the solid
region in the rst octant bounded above by the sphere = 3 and below by the cone
= 6.
6 2 3 2 6 2 3 2
0 0 0
sin = 0
sin 0 0

6 2 1 3 3
= cos 0 0 3 0

3 9
= 1 (9) = 2 3
2 2 4

19. The solid is most conveniently described if we use cylindrical coordinates:
= ( )|0 2
0 3 0 2 . Then
2 3 2
( ) = 0 0 0
( cos sin ) .

21. In spherical coordinates, is represented by {( )|0 5 0 2 0 }. Thus
2 2 2 2 2 5 2 2 2 2 5 6
( + + ) = 0 0 0
( ) sin = 0
sin 0 0
2 1 7 5 78,125
= cos 0 0 7 0
= (2)(2 ) 7

312,500
= 7
140,249 7

23. In spherical coordinates, is represented by ( ) 1 2 0 2
0 2
. Thus
2 2 2 2 2 2 2 3
= 0 0 1
( cos ) sin = 0
cos sin 0 1

2 2 1 4 2
= 1
2
sin2 0 0 4 1
= 1
2 2
15
4
= ...

Archive contentsContenu de l'archive

Action(s) SizeTaille FileFichier
3.06 Ko KB readme.txt
3.65 Ko KB lisezmoi.txt
1.17 Ko KB testehpp.hpprgm
37.10 Ko KB testehpp.hpappdir.zip
95 octets bytes appslist.txt

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