Find the area of the helicoid (or spiral ramp) with vector equation r(u, v) = ucos(v) i + usin(v) j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 9π.

The area f the helicoid ramp  is:

∫∫A) | r(u) *r( v) | dudv

The solution is:

A = 10.35×π  square units

r ( u , v ) = u×cosv i +  u×sinv j +v k

To get

r (u ) = δ(r ( u , v ) ) / δu   = [ cosv , sinv , 0 ]

r ( v ) = δ(r ( u , v ) ) / δv  = [ -u×sinv , u×cosv , 1 ]

The vectorial product is:

                                 i                    j              k

r (u ) * r ( v )            cosv             sinv           0

                            -u×sinv           u×cosv       1

r (u ) * r ( v )  =  i × ( sinv - 0 )  - j × ( cosv - 0 ) + k ( u×cos²v + u× sin²v )

r (u ) * r ( v )  = sinv i - cosv j + u k

Now

| r (u ) * r ( v ) |  = √sin²v + cos²v + u²  = √ 1 + u²

Then  

A = ∫₀ (9π) dv    ∫₀¹  √ 1 + u²  du    

 ∫₀¹  √ 1 + u²  du    = 1.15  

A = 1.15 × v |( 0  , 9π )

A = 10.35×π  square units

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