Evaluate the integral over the helicoid [Surface integrals]

In summary, the integral \int\int_S \sqrt{1+x^{2}+y^{2}}dS where S:{ r(u,v) = (ucos(v),usin(v),v) | 0\leq u\leq 4,0\leq v\leq 4\pi } can be evaluated by first substituting x=ucos(v) and y=usin(v) and finding the jacobian determinant to be u. This results in the integral \int_0^{4\pi} \int_0^4 u\sqrt{(1+u^{2})} dudv. After a simple substitution, the integral can be simplified to 2\
  • #1
ysebastien
6
0

Homework Statement


Evaluate the integral [itex]\int\int_S \sqrt{1+x^{2}+y^{2}}dS[/itex] where S:{ r(u,v) = (ucos(v),usin(v),v) |[itex] 0\leq u\leq 4,0\leq v\leq 4\pi[/itex] }


2. The attempt at a solution

Here is my attempt, I am fairly sure I am right, but it is an online assignment and it keeps telling me I am wrong. I just wanted to double check before I contact the professor to see if he made a mistake.

Let x=ucos(v),y=usin(v) and,

the jacobian determinant is [itex]u(cos^{2}(v)+sin^{2}(v))=u[/itex]

now my new integral is

[itex]\int_0^{4\pi} \int_0^4 u\sqrt{(1+u^{2})} dudv[/itex]

Now this is a fairly straightforward problem and I do a simple substitution to get

let [itex]\phi=1+u^{2},du=2u[/itex]

[itex]\frac{1}{2}\int_0^{4\pi} \int \sqrt{\phi} d\phi dv=2\pi[\frac{2}{3}\phi^{\frac{3}{2}}]=2\pi[\frac{2}{3}(1+u^{2})^{\frac{3}{2}}]_0^4[/itex]

and finally after plugging in the values, I get

[itex]2\pi(\frac{2}{3} 17^{\frac{3}{2}} - \frac{2}{3})[/itex]

Does anyone else see a flaw in this? again, I am pretty sure I am right but would appreciate it immensely if someone could point out my mistake!

Thank you

EDIT: Also if I made any typos in the equations my apologies, this is my first time editing with latex commands
 
Last edited:
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  • #2
Nevermind.
 

1. What is a helicoid?

A helicoid is a three-dimensional surface that resembles a spiral staircase. It is formed by a ruled surface in which the straight lines making up the surface are all parallel to a single axis and wrap around the surface in a helical pattern.

2. What is a surface integral?

A surface integral is a type of multiple integral used to calculate the total value of a function over a two-dimensional surface. It is similar to a regular integral, but instead of integrating over a one-dimensional interval, it integrates over a two-dimensional surface.

3. How do you evaluate an integral over a helicoid?

To evaluate an integral over a helicoid, you first need to parameterize the surface by defining two variables, usually denoted as u and v, that represent the coordinates on the surface. Then, you can use the surface area element formula and the function you are integrating to set up the integral and solve for the final value.

4. What is the surface area element formula?

The surface area element formula is a mathematical equation used to calculate the surface area of a parametric surface. It is typically written as dS = √(1 + (du/dv)^2 + (du/dv)^2) dA, where du and dv are the partial derivatives of the parameterization variables u and v, and dA is the area element in the u-v plane.

5. What are some applications of surface integrals?

Surface integrals have many applications in physics, engineering, and mathematics. They are used to calculate the flux of a vector field through a surface, the mass and center of mass of a three-dimensional object, and the work done by a force on a moving object. They are also used in various areas of science, such as fluid dynamics, electromagnetism, and heat transfer.

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