Great Circles, the key to competent navigation! Passing through opposite points of the surface. It don't mean a thing, if it got that spin...because there is some interesting physics that happen on the surface of a rotating sphere.
You are marvelous sir. Just like Ramanujan. I like the way you teach, its fabulous. I want to know where did you study from and in which country do you teach mathematics, and importantly are you really a school teacher? Hope you will reply. Thank you
You've made it not only understandable but enjoyable too! Thank you, Sir!!
can't thank you enough!! no every teacher is as competent in teaching like you. love it!!!
finally someone who explalins how the formula makes sense. Thanks a lot sir!
Now I understand, thank you! (My teacher never explains in detail like you do)
Great Circles, the key to competent navigation! Passing through opposite points of the surface. It don't mean a thing, if it got that spin...because there is some interesting physics that happen on the surface of a rotating sphere.
He's the best!!!
You are marvelous sir. Just like Ramanujan. I like the way you teach, its fabulous. I want to know where did you study from and in which country do you teach mathematics, and importantly are you really a school teacher?
Hope you will reply.
Thank you
Mohit Jassi
He really is a teacher here in a local school in Sydney Australia.
When looked to an hemisphere on top you can see it like a circle so it must be 2πr^2 a sphere`s surface area. Isnt it?
Don't know why I am replying ,you need to proof it using Conformal map(I think)
Hello , please can you explain how to convert the surface area of the sphere to the area of rectangular
Conformal map
Cylinder first, then turn it into a rectangle.
Hey what's wrong with the antarctic circle? You're an Australian!
Lucky students who could sit in his classes
coool
you are the marker walter lewin
🙏🙏🙏🙏🙏
2:50 theory and practice are two completely different things :) :)
So why don't you simplify it? Use the diameter. A= pi times D squared. Much easier.
The radius is more used for example area of a circle = pi*r^2 and area of circle = 2pi*r
Using the integral by thinking, it's easier:
surface = 2 * integral { [ 2 * pi * (R^2 - z^2)^0.5 ] dz | (0, R) }
Due to abandoned tiny curved areas, the real magnitude must be MORE than "12.56".
But it is extremely small...so makes no sense at all
hola
2nd comment