Question

math problem

Answer 1

Try using logarithms!

Answer 2

what's p(r) when r=R

Answer 3

Use integration

Answer 4

i gotta pass in this .... or at least on giving an answer since i gots no idea where id start

Answer 5

no, I just need the function value of p(r) when r=R

Answer 6

it doesn't look well defined at r=R

Answer 7

b/c the way problem is stated both function works when r=R

Answer 8

\[\int_0^R \rho_0(\frac{1-4r}{3R})dr\]

Answer 9

no, agdgdgdgwng, that wouldn't work

Answer 10

:-( but I did physics problems like this 4-8 months ago

Answer 11

well i probably shouldn't butt in but it sure looks like you have it defined at r = R in two different ways

Answer 12

yeah, satelllite , that's what troubling me

Answer 13

http://www.cramster.com/answers-jan-10/physics/gausss-lawplease-nonuniform-spherically-symmetric-distribution-cha_748512.aspx?rec=0

Answer 14

they want me to pay to see the answer

Answer 15

maybe it is a typo?

Answer 16

that is what I am thinking

Answer 17

since you need to integrate the mixup at one value shouldn't change anything (set of measure zero)

Answer 18

i can view the answer. let me take a screen shot and send it

Answer 19

thanks

Answer 20

yay thanks

Answer 21

hold on one second or two

Answer 22

I really should practice my physics so I won't fail when I finally get into a university.

Answer 23

Until I'm confident with my ability to solve physics problems.

Answer 24

ok really feel like a moron now. i took 3 screen shots but cannot find them! where are they saved in ubuntu??

Answer 25

lol

Answer 26

/dev/null maybe

Answer 27

If all else fails, try typing the solution with your 1337 LaTeX skills.

Answer 28

that was strange. after 5 tries i got option to save...

Answer 29

can you read that one?

Answer 30

I can read them both, thanks

Answer 31

whew. sorry it took a while

Answer 32

I don't like how they used the variable r as a limit of integration and as a variable they were integrating with respect to.

Answer 33

we were integrating
\[\int_0^R \rho \left(4 \pi r^2\right) \left(1-\frac{4 r}{3 R}\right) \, dr\]

we should have been doing

\[\int_0^r \rho \left(4 \pi r^2\right) \left(1-\frac{4 r}{3 R}\right) \, dr\]

Answer 34

that's why I was hesitant to use r as limit when it is also a variable

Answer 35

it look like the charge is 0 when r=R

Answer 36

*looks

Answer 37

\[\int_0^r \rho \left(4 \pi r^2\right) \left(1-\frac{4 r}{3 R}\right) \, dr\]=

\[\frac{4}{3} \pi r^3 \rho -\frac{4 \pi r^4 \rho }{3 R}\]

now it checks out

Answer 38

which is the same thing we had yesterday if you did this...
\[\int _0^{2 \pi }\int _0^{\pi }\int _0^r\rho_0\gamma ^2 \left(1-\frac{4 \gamma }{3 R}\right)\text{ }\sin (\phi )d\gamma d\phi d\theta\]

Answer 39

yeah, nice

Answer 40

But I guess not many people remember spherical coordinate integration out of Calc III, lol

Answer 41

i guess not ;)

Answer 42

I have to be ready for all the surface integral problem for next chapter

Answer 43

sounds like fun

Answer 44

It is , when I get a right answer, lol

Answer 45

Thanks everyone