How to solve a surface integral problem involving a helicoid?

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SUMMARY

The discussion focuses on evaluating the surface integral ∫∫S √(1 + x^2 + y^2) dS over the helicoid defined by r(u, v) = ucos(v)i + usin(v)j + vk, with the limits 0 ≤ u ≤ 4 and 0 ≤ v ≤ 3π. The initial attempt incorrectly simplified the integrand to √(1 + u) and miscalculated the length of the cross product, leading to an erroneous integral setup. The correct approach involves calculating the cross product of the surface parameterization, resulting in a length of √(1 + u^2), and thus the integral should be set up as ∫∫(1 + u^2) dudv with the specified limits.

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benji55545
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Homework Statement



Evaluate ∫∫S √(1 + x^2 + y^2) dS where S is the helicoid: r(u, v) = ucos(v)i + usin(v)j + vk, with 0 ≤ u ≤ 4, 0 ≤ v ≤ 3π


The Attempt at a Solution



What I tried to do was say x=ucos(v) and y=usin(v), then I plugged those into the sqrt(1+x^2+y^2) eq, which I ended up simplifying to sqrt(1+u) somehow. I then took the cross product of the surface eq differentiated (respect to u) with eq (respect to v). I found the length of that, which I somehow got to be sqrt(2).

I then said that the intergal was ∫∫sqrt(1+u)*sqrt(2)du*dv
where the limits of integration are the limits of u and v given above.
Well this is wrong. Hah!

Any help would be appreciated.
 
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benji55545 said:

Homework Statement



Evaluate ∫∫S √(1 + x^2 + y^2) dS where S is the helicoid: r(u, v) = ucos(v)i + usin(v)j + vk, with 0 ≤ u ≤ 4, 0 ≤ v ≤ 3π


The Attempt at a Solution



What I tried to do was say x=ucos(v) and y=usin(v), then I plugged those into the sqrt(1+x^2+y^2) eq, which I ended up simplifying to sqrt(1+u) somehow. I then took the cross product of the surface eq differentiated (respect to u) with eq (respect to v). I found the length of that, which I somehow got to be sqrt(2)..
I'm not happy with the "somehow"! That's where you went wrong. The cross product you speak of (also called the 'fundamental vector product') is [itex]sin(v)\vec{i}- cos(v)\vec{j}+ u\vec{k}[/itex]
(did you forget the "u" multiplying -cos(v) and sin(v) in the derivative with respect to v?) and that has length [itex]\sqrt{1+ u^2}[/itex]. Thus your integral is
[tex]\int_{u= 0}^4\int_{v=0}^{2\pi}(1+ u^2)dudv[/tex]

I then said that the intergal was ∫∫sqrt(1+u)*sqrt(2)du*dv
where the limits of integration are the limits of u and v given above.
Well this is wrong. Hah!

Any help would be appreciated.
 
Last edited by a moderator:
Thanks for the reply.
I still cannot seem to get the correct answer. I solved the double integral you provided and was incorrect, then changed the limit on the integral with respect to u to 3pi like the limits given, but still no luck. Any suggestions on what I'm doing wrong? Thanks.
 
Last edited:

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