Finding Surface Area Cone through integration

  • Thread starter maupassant
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  • #1
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Homework Statement



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!



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!

Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
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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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