How to solve a surface integral problem involving a helicoid?

In summary, the problem involves evaluating the integral of √(1 + x^2 + y^2) over a helicoid surface given by r(u, v) = ucos(v)i + usin(v)j + vk, with limits 0 ≤ u ≤ 4 and 0 ≤ v ≤ 3π. After setting x=ucos(v) and y=usin(v) and simplifying the integrand to √(1+u), the attempt at a solution involved taking the cross product of the surface equation differentiated with respect to u and v and finding its length to be √(1+u^2). However, there was an error in the calculation and the correct integral should be
  • #1
benji55545
11
0

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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  • #2
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:
  • #3
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:

1. What is a surface integral problem?

A surface integral problem is a mathematical problem that involves calculating a value over a given surface. It is similar to a regular integral, but instead of integrating over a one-dimensional line, it integrates over a two-dimensional surface.

2. How is a surface integral problem different from a regular integral?

A surface integral problem is different from a regular integral in that it involves integrating over a two-dimensional surface instead of a one-dimensional line. This means that the limits of integration are given by the boundaries of the surface, rather than by a single number.

3. What are some real-world applications of surface integrals?

Surface integrals have various applications in physics and engineering, such as in calculating the flux of a vector field through a surface, determining the mass and center of mass of a 3D object, and finding the work done by a force on a surface. They are also used in fluid mechanics, electromagnetism, and computer graphics.

4. What are the different types of surface integrals?

There are two main types of surface integrals: the surface integral of a function and the surface integral of a vector field. The former involves integrating a scalar-valued function over a surface, while the latter involves integrating a vector field over a surface to calculate the flux or circulation.

5. How can I solve a surface integral problem?

Solving a surface integral problem involves first determining the limits of integration, either by parametrizing the surface or using a given parametric equation. Then, the integrand needs to be set up according to the type of surface integral being calculated. Finally, the integral can be evaluated using appropriate integration techniques, such as substitution or double/triple integration.

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