# Surface area of a cone problem

## Homework Statement

The question is to derive the surface area of a cone.

## Homework Equations

slant= square root ( r^2 + h^2)
surface area= int int [square root(fx^2 + fy^2 +1) da]
surface area of cone side= pi *r(r^2+h^2)
3d cone formula: z= h/r(squareroot x^2+y^2)

## The Attempt at a Solution

by looking at the structure I know that it is the area of the base (circle) + the area of the slant/side, but when I solve for the surface area using double integrals I'm stuck w/ squareroot 2 in the formula. How can I cancel that out?

i calculated fx as hx/rsqareroot(x^2+y^2)
and fy as hy/rsquareroot(x^2+y^2)

i plugged that into the formula for surface area and got: int int [h/r squareroot(2)] r dr d@
it feels like it isn't right and I don't know how to cancel the sqareroot(2) during integration. Can someone hint me in the right direction?

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Dick
Homework Helper
fx and fy look ok. But when I compute sqrt(1+(fx)^2+(fy)^2) I don't get what you got for the integrand (in particular, no sqrt(2)). Can you tell us how you got that?

fx and fy look ok. But when I compute sqrt(1+(fx)^2+(fy)^2) I don't get what you got for the integrand (in particular, no sqrt(2)). Can you tell us how you got that?
sure:
int int squareroot [(hx/rsquareroot(x^2+y^2))^2 + (hy/rsquareroot(x^2+y^2))^2 + 1] da
int int squareroot [(h^2*x^2/r^2 (x^2+y^2)) + (h^2*y^2/r^2(x^2+y^2)) + 1] da
int int squareroot [(h^2/r^2) x^2/(x^2+y^2) + y^2/(x^2+y^2) +1] da
the +1 should change into: (x^2+y^2)/(x^2+y^2)
int int squareroot [h^2/r^2 (x^2+ y^2/(x^2+y^2)) + x^2+y^2/(x^2+y^2)] da
int int h/r squareroot(2) r dr d@

does it look right? :/ theres probablt something big that I'm missing but its so hard to see

Dick