# Finding Surface Area Cone through integration

1. Oct 2, 2011

### maupassant

1. The problem statement, all variables and given/known data

Find the surface area of the cone with the following equations:
x= u sin(a)cos(v) , y= u sin(a)sin(v), z=u cos(a)
where 0<=u <=b , 0<=v<=2(pi), a is constant!

3. The attempt at a solution

Trying to solve this I first calculate the absolute value of the cross product of r'(u) and r'(v):

ABS(r'(u) x r'(v)) = SQRT(u^2 (sin(a) ^2)) = u sin(a)

Then I try to integrate this result by calculating

∫ ∫ (u sin(a)) du dv = ∫ (1/2)(b^2) (sin(a)) dv = (pi) (b^2) sin(a)

with 0<=u<=b and 0<=v<=2(pi) as limits of integration.

This result however does not correspond with the formula to find the surface area of a cone.
Could someone help out with this problem?
Thank you!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 2, 2011
2. Oct 3, 2011

### Eynstone

Let a surface be X=(x(u,v),y(u,v),z(u,v)) . The element of area will be the cross product of Xu & Xv ( Xu is the partial derivative with respect to u &c.).Integrate this within the limits.