1. ## Quick Integration Check please!

Hi can you just check that I have this integration correct?

(1 - r^2/R)

Solution

( r - r^3/3)

Cheers!

2. What is R and how is it related to r? Convention dictates that variables are case sensitive

$\displaystyle \int 1- \dfrac{r^2}{r} dr = \int (1-r) dr$

If R is some constant:

$\displaystyle \int 1- \dfrac{r^2}{R} dr = \int 1 - ar^2$ where $a = \dfrac{1}{R}$

edit: you can check integration by differentiating the result and since $\dfrac{d}{dr} \left(r - \dfrac{r^3}{3}\right) = 1-r^2$ it's not correct

3. I would get a different answer to yours but I'm not certain what R is....

4. Sorry Ill explain myself better. The question relates to charge distribution that is spherically symmetrical. This charge distribution is over the range:

A(1 - r/R) for r< R

and

0 for r > R

Where A is a constant

Making a few assumptions about the charge distribution we get:

Q = A( 1 - r/R) × 4πr^2* δr = 4πA(1 - r^3/R)δr

π = pi

So to find the total charge over the whole of sphere we take integral from R to 0:

Q = 4πA(1 - r^3/R) = 4πA [This is the bit i am obviously getting wrong!] = ??????

Sorry about using the correct symbols but i don't seem to be able to cut and paste from word into the forum