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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.
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