Question

Help with definite integral.

Answer 1

\[\frac{ \sigma R^2 }{ 4 \epsilon_{0} }\int\limits_{\infty}^{h}dh'\frac{ h' }{ \sqrt{(R^2+h'^2)^3} }=\frac{\sigma R}{16 \epsilon_{0}}\left( \cos[2 \arctan h]+1 \right)\]Whats on the right is the result I got, I took the R out of the square root and caled h/r=tg u. I'd like to know If thats the right result when doing this way or not, couse I can't find any mistake, but in the test it said it was wrong.

Answer 2

@TuringTest

Answer 3

let us call h'^2 + R^2 = u
then du = 2h'dh
so du/2 = h'dh

and we have

integral of du / 2sqrt(u^3) ..

Answer 4

I know it works better this way, I saw it later but that was not my question, I need to know if calculating it by the way I did in the test it gives what I got.

Answer 5

ok lets do it together
h/r = tgu
so h = rtgu
dh = rsec^2(u)du

so your integral becomes
r^2sec^2(u)tg(u)du / r^3sec^3(u) =
tg(u)du/rsec(u) =
sin(u)du/r =
-cos(u)/r ?

Answer 6

maybe ive missed something

Answer 7

no .. what i did is fine
we can see that it is the same as the other way :

in the first method (the easy one) we get -1/sqrt(r^2+h^2)

in the other we got -cos(u)/r
now we know tan(u) = h/r
so it means cos(u) = r/sqrt(r^2+h^2)
plug it into -cos(u)/r
give us -1/sqrt(r^2+h^2)
so same..

Answer 8

Ohh I got it, after the first transformation in the denominator I put r not r^3, but but that only affects the final result, when cheking the secs out I also messed up.
Thank you, I was not seing that.

Answer 9

yw