# Surface Integral of a helicoid.

Gold Member

## Homework Statement

Evaluate

$\int\int_{S}\sqrt{1+x^2+y^2} dS$

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

$\int_{0}^{4\pi}\int_{0}^{2} \sqrt{1+u^2}u du dv$

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

$\frac{4\pi}{3}(5\sqrt{5}-1)$

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

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper

## Homework Statement

Evaluate

$\int\int_{S}\sqrt{1+x^2+y^2} dS$

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

$\int_{0}^{4\pi}\int_{0}^{2} \sqrt{1+u^2}u du dv$

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

$\frac{4\pi}{3}(5\sqrt{5}-1)$

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?

Gold Member
S is a function of u, and v. The u du comes from the u-substitution.

Sorry
$\frac{1}{2}\int_{0}^{4\pi}\int_{1}^5 \sqrt{w} dw dv$

Is the after the u-sub I think.

Dick
$\frac{1}{2}\int_{0}^{4\pi}\int_{1}^5 \sqrt{w} dw dv$