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

[nhrock3]

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.)

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

$$\rho\le 2\cos \phi$$

??

 

[tiny-tim]

hi nhrock3!

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 ?

 

[nhrock3]

hi nhrock3!

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 ?

OP^2=0C^+PC^2
how it changes the intervals?

 

[tiny-tim]

OP^2=0C^+PC^2

no, it's not a right-angle

 

[SammyS]

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.

 

[nhrock3]

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

 

[tiny-tim]

C is the centre, so CO = CP = r = 1.

Angle COP = φ.

So how long is OP?​

 

[micromass]

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.

 

[nhrock3]

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?

 

[tiny-tim]

OP^2=CO^2+CP^2-2OC*PCcos(180-2φ)

what to do now?

well, now put the numbers in

 

[nhrock3]

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??

 

[SammyS]

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

 

[tiny-tim]

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 )

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

(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

in other words, put CO = CP= PC = 1 …

what do you get?​