• Support PF! Buy your school textbooks, materials and every day products Here!

Surface Integral of a helicoid.

  • #1
ElijahRockers
Gold Member
270
10

Homework Statement



Evaluate

[itex]\int\int_{S}\sqrt{1+x^2+y^2} dS[/itex]

S is the helicoid with vector equation r(u,v) = <u cos(v), u sin(v), v>

0<u<2, 0<v<4pi

The Attempt at a Solution



If I replace the term under the radical with its vector equation counterpart, and multiply that by the cross product of the partials of r(u,v) with respect to u and v, i get

[itex]\int_{0}^{4\pi}\int_{0}^{2} \sqrt{1+u^2}u du dv[/itex]

From there I can do a u-substitution (ill just call it a ω-sub so as not to confuse) with ω=1+u2, and dω/2 = udu.

When I work this out, I get

[itex]\frac{4\pi}{3}(5\sqrt{5}-1)[/itex]

But according to the software this answer is incorrect. Anyone notice a mistake?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement



Evaluate

[itex]\int\int_{S}\sqrt{1+x^2+y^2} dS[/itex]

S is the helicoid with vector equation r(u,v) = <u cos(v), u sin(v), v>

0<u<2, 0<v<4pi

The Attempt at a Solution



If I replace the term under the radical with its vector equation counterpart, and multiply that by the cross product of the partials of r(u,v) with respect to u and v, i get

[itex]\int_{0}^{4\pi}\int_{0}^{2} \sqrt{1+u^2}u du dv[/itex]

From there I can do a u-substitution (ill just call it a ω-sub so as not to confuse) with ω=1+u2, and dω/2 = udu.

When I work this out, I get

[itex]\frac{4\pi}{3}(5\sqrt{5}-1)[/itex]

But according to the software this answer is incorrect. Anyone notice a mistake?
You seem to have turned dS into ududv. How did you do that?
 
  • #3
ElijahRockers
Gold Member
270
10
S is a function of u, and v. The u du comes from the u-substitution.

Sorry
[itex]\frac{1}{2}\int_{0}^{4\pi}\int_{1}^5 \sqrt{w} dw dv[/itex]

Is the after the u-sub I think.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
S is a function of u, and v. The u du comes from the u-substitution.

Sorry
[itex]\frac{1}{2}\int_{0}^{4\pi}\int_{1}^5 \sqrt{w} dw dv[/itex]

Is the after the u-sub I think.
I'm asking what you got for dS using the cross product.
 

Related Threads for: Surface Integral of a helicoid.

Replies
1
Views
3K
  • Last Post
Replies
1
Views
2K
Top