Evaluate the surface integral. s y ds, s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v ≤ π.

The surface integral (y ds), s is the helicoid with vector equation r(u,v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v ≤ π is and this can be determine by doing the integration.

Given :

The surface integral (y ds), s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v ≤ π.

The following calculation can be used in order to evaluate the surface integral.

The surface integral is given by:

 --- (1)

where and is given by:

Now, evaluate .

Now, substitute the known values in the expression (1).

Now, integrate the above expression.

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Compute the surface element:

So the integral is

b: |12 – x| ≤ 0.25

step-by-step explanation:      

let x be the possible lengths in inches, of the ribbons jaime cut.

we have been given that jaime is cutting strips of ribbon for a project. he tries to cut each piece to be 12 inches long, but sometimes he cuts it too long or too short by as much as 0.25 inch.  

we can represent this information in an inequality as:

by the definition of absolute value function we will get,

upon combining our inequalities we will get,

therefore, the inequality   describes all the possible lengths (x), in inches, of the ribbons jaime cut.

step-by-step explanation: