What is the mistake here

  • Thread starter nhrock3
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  • #1
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why does the prof wrote a mistake there and wrote 2cos theta in there
??

(i didnt typed the question correctly,the original is
i need to calculate the integral on the volume enclosed by z>=0 and the sphere wich is written in the photo.)
24oq2ww.jpg

how mathematicky can i get this transition

how to get this expression
from cartesian to polar and get the 2 cos teta on the interval

[TEX]\rho\le 2\cos \phi[/TEX]

??
 

Answers and Replies

  • #2
tiny-tim
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hi nhrock3! :smile:

if the origin is O, and if the centre of the sphere is C, and if P is a point on the sphere,

what are the angles of triangle OCP, and what is the length of OP ? :wink:
 
  • #3
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hi nhrock3! :smile:

if the origin is O, and if the centre of the sphere is C, and if P is a point on the sphere,

what are the angles of triangle OCP, and what is the length of OP ? :wink:
OP^2=0C^+PC^2
how it changes the intervals?
 
  • #4
tiny-tim
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  • #5
SammyS
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What are the lengths of OP & PC ?

Isn't φ the angle from the z axis to OP ?

Use the law of cosines to answer tiny-tim's question.
 
  • #6
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What are the lengths of OP & PC ?

Isn't φ the angle from the z axis to OP ?

Use the law of cosines to answer tiny-tim's question.

i cant imagine the angle


Isn't φ the angle from the z axis to OP ?


could you draw it please
 
  • #7
tiny-tim
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C is the centre, so CO = CP = r = 1.

Angle COP = φ.

So how long is OP?​
 
  • #8
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Or just use that [itex]x^2+y^2+z^2=\rho^2[/itex] and [itex]z=\rho \cos\phi[/itex] to transform

[tex]x^2+y^2+z^2=2z[/tex]

into an equation that uses only rho and phi.
 
  • #9
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C is the centre, so CO = CP = r = 1.

Angle COP = φ.

So how long is OP?​
OP^2=CO^2+CP^2-2OC*PCcos(180-2φ)

what to do now?
 
  • #10
tiny-tim
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OP^2=CO^2+CP^2-2OC*PCcos(180-2φ)

what to do now?
well, now put the numbers in :confused:
 
  • #11
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ok i got it
if we try and solve it in a different way.
but if we do variable change
x=u y=v z-1=w
then the jacobian is 1
and the integral is a ball of radius from 0 to 1

unlike here where our radius is from 0 to 2cos

why??
 
  • #12
SammyS
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Placing the origin at the center of the sphere naturally makes the limits of integration simple for a sphere. I hope you changed the integrand accordingly.
z → w+1
 
  • #13
tiny-tim
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ok i got it
if we try and solve it in a different way.
no, nhrock3, you haven't got it …

you still haven't a clue why your prof did that …
why does the prof wrote a mistake there and wrote 2cos theta in there
your prof is a clever guy who knows how best to teach this subject (and who knows what's coming up in the exams :rolleyes:)

it is not clever for you to give up on his method and to "try and solve it in a different way" :redface:

(and presumably you still think your prof wrote a mistake?)

start again …​
OP^2=CO^2+CP^2-2OC*PCcos(180-2φ)

what to do now?
well, now put the numbers in :confused:
in other words, put CO = CP= PC = 1 …

what do you get?
 

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