A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Hi
A question?
anonymous
 4 years ago
Hi A question?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can explain how this two sides are equal?\[\cos \theta da=r^2d \Omega\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where da is surface element in spherical coordinate & omeg is solid angle

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0It looks wrong. If you integrate over the entire sphere, the right hand side is \[ 4 \pi r^2 \] but the left hand side is \[ 8 \pi r^2 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's in classical electrodynamics (by Jackson)page28

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0I think this is the same formula we are finding in figure 4.2.4 on this page http://ocw.mit.edu/courses/physics/802scphysicsiielectricityandmagnetismfall2010/conductorsandinsulatorsconductorsasshields/MIT8_02SC_notes9.pdf though the area is not represented as a differential here. I trust that James' math is right though, so that must have some change in the problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0r square* d(sigma) = r*d(theta) *r*d(phi) =projection of area element=dA cos(theta) ....in azimuthal coordinates..dw:1334694350771:dw
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.