Ya, I got that a couple times too (after I remembered the orientation), however the answer I was provided is 7/6 pi. That could be wrong. I'll re-do the surface integral the other way and see what I come up with I guess.
I was told this problem could be done with divergence theorem, instead of as a surface integral, by adding the unit disc on the bottom, doing the calculation, then subtracting it again.
The Attempt at a Solution
for del . f I get i + j =...
You are correct in both cases :)
B is pointing into the screen, electrons just follow the simple right hand rule, and will be deflected in the direction that your fingers curl. Flip this 'normal' alignment to deal with positive charges.
Your logic on the second is just fine. The equation for...
Yes, the amplitude in this case is arbitrary, it just equals A. In other words if you start a pendulum swinging on Earth by pulling it to A, then go to the moon and start it again by pulling it to A, the difference is in the period (which gets longer by the relationship above).
If you do the...
Pressure in this case doesn't depend on volume, it just depends on gravity, density and the height of the fluid column above you. You've probably seen the equation before- p=d*g*h.
The total, then, is just that plus atmospheric pressure.
I substitute sin(theta) for d(omega), and the integral for 1/sin(theta) is log(tan(theta/2)) at least as given by mathematica and the back of my book.
I kind of figured I shouldn't be getting that, which is what makes me think I'm approaching the problem all out of whack.
At least it should be a simple integral...
The whole text is here- http://i35.tinypic.com/2nisnp.jpg
Basically (I think) I need to integrate the differential over all angles theta and phi, and get sigma(naught) back out.
given in pic...