The potential at points around a cone

[Identity]

Homework Statement

"A conical surface (an empty ice-cream cone) carries a uniform surface charge \sigma. The height of the cone is h, as is the radius of the top. Find the potential at the centre of the top, taking infinity as reference point." - Griffiths

My result for the potential differs from the answer's. Can someone please check my solution

Homework Equations

V = \iint_S \frac{1}{4\pi \epsilon_0}\frac{\sigma}{x} dS, where x is the distance from the source to the point.

The Attempt at a Solution

My diagram is in the attached picture.

First off, I used cylindrical coordinates with the equation z = r, z > 0 to graph the cone
I found dS = \sqrt{2}zdA

The distance from any point on the cone to the origin is \sqrt{2} z, so using the cosine law,
x = \sqrt{2z^2-2hz+h^2}

So we have

V = \int_0^h\int_0^{2\pi} \frac{1}{4\pi \epsilon_0} \frac{\sigma}{\sqrt{2z^2-2hz+h^2}} \sqrt{2}z d\theta dz=\frac{\sigma h}{4\epsilon}\ln(2\sqrt{2}+3)


Answer: V = \frac{\sigma h}{2\epsilon}\ln(1+\sqrt{2})


thanks!

 

[tiny-tim]

V = \int_0^h\int_0^{2\pi} \frac{1}{4\pi \epsilon_0} \frac{\sigma}{\sqrt{2z^2-2hz+h^2}} \sqrt{2}z d\theta dz=\frac{\sigma h}{4\epsilon}\ln(2\sqrt{2}+3)


Answer: V = \frac{\sigma h}{2\epsilon}\ln(1+\sqrt{2})

Hi Identity!

erm … 2ln(1 + √2) = ln(3 + 2√2)

 

[Identity]

oh my!
thankyou tiny-tim