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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a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|

it's possible to build a triangle with side lengths of 3, 3, and 9.- true ( b.)