Surface Integral of a helicoid.

[ElijahRockers]

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?

 

[Dick]

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?

 

[ElijahRockers]

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.

 

[Dick]

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.