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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

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