1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding Surface Area Cone through integration

  1. Oct 2, 2011 #1
    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. jcsd
  3. Oct 3, 2011 #2
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook