Question

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.

Answer 1

Let $H$ denote the helicoid parameterized by

$\mathbb r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+v\,\mathbf k$

for $0\le u\le1$ and $0\le v\le9\pi$. The surface area is given by the surface integral,

$\displaystyle\iint_H\mathrm dS=\iint_H\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$

We have

$\mathbf r_u=\dfrac{\partial\mathbf r(u,v)}{\partial u}=\cos v\,\mathbf i+\sin v\,\mathbf j$
$\mathbf r_v=\dfrac{\partial\mathbf r(u,v)}{\partial v}=-u\sin v\,\mathbf i+u\cos v\,\mathbf j+\mathbf k$
$\implies\mathbf r_u\times\mathbf r_v=\sin v\,\mathbf i-\cos v\,\mathbf j+u\,\mathbf k$
$\implies\|\mathbf r_u\times\mathbf r_v\|=\sqrt{1+u^2}$

So the area of $H$ is

$\displaystyle\iint_H\mathrm dS=\int_{v=0}^{v=9\pi}\int_{u=0}^{u=1}\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$
$=\dfrac{9(\sqrt2+\sinh^{-1}(1))\pi}2$

Answer 2

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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