Hi can you just check that I have this integration correct?
(1 - r^2/R)
Solution
( r - r^3/3)
Cheers!
What is R and how is it related to r? Convention dictates that variables are case sensitive
$\displaystyle \displaystyle \int 1- \dfrac{r^2}{r} dr = \int (1-r) dr$
If R is some constant:
$\displaystyle \displaystyle \int 1- \dfrac{r^2}{R} dr = \int 1 - ar^2$ where $\displaystyle a = \dfrac{1}{R}$
edit: you can check integration by differentiating the result and since $\displaystyle \dfrac{d}{dr} \left(r - \dfrac{r^3}{3}\right) = 1-r^2$ it's not correct
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