Why is the professor's method for solving the integral more efficient?

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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 which 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

\rho\le 2\cos \phi

??
 
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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:
 
tiny-tim said:
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?
 
nhrock3 said:
OP^2=0C^+PC^2

no, it's not a right-angle
 
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.
 
SammyS said:
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 can't imagine the angle


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


could you draw it please
 
C is the centre, so CO = CP = r = 1.

Angle COP = φ.

So how long is OP?​
 
Or just use that x^2+y^2+z^2=\rho^2 and z=\rho \cos\phi to transform

x^2+y^2+z^2=2z

into an equation that uses only rho and phi.
 
tiny-tim said:
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
nhrock3 said:
OP^2=CO^2+CP^2-2OC*PCcos(180-2φ)

what to do now?

well, now put the numbers in :confused:
 
  • #11
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
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
nhrock3 said:
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 …
nhrock3 said:
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 …​
nhrock3 said:
OP^2=CO^2+CP^2-2OC*PCcos(180-2φ)

what to do now?
tiny-tim said:
well, now put the numbers in :confused:

in other words, put CO = CP= PC = 1 …

what do you get?
 
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