Question

Evaluate iint S sqrt 1+x^ 2 +y^ 2 dS where S is the helicoid: r(u, v) = u * cos (v) * i + u * sin (v) * j + vk with 0 <= u <= 1, 0 <= v <= 2pi

Answer 1

Given the parameterization

`\vec r(u,v) = u\cos(v) \, \vec\imath + u\sin(v) \,\vec\jmath + v \,\vec k`

take the normal vector to be

`\vec n = (\partial\vec r)/(\partial u) * (\partial \vec r)/(\partial v) = \sin(v) \,\vec\imath - \cos(v) \,\vec\jmath - u \,\vec k`

(The order of partial derivatives in the cross product doesn't matter since this a scalar line integral.)

Compute the magnitude of the normal vector.

`\|\vec n\| = √(\sin^2(v) + (-\cos(v))^2 + (-u)^2) = √(1+u^2)`

so that the area element reduces to

`dS = \|\vec n\| \, du\,du = √(1+u^2)\,du\,du`

Evaluate the integrand at
`\vec r` to get

`√(1 + (u\cos(v))^2 + (u\sin(v))^2) = √(1 + u^2)`

The surface integral reduces to

`\displaystyle \iint_S √(1+x^2+y^2) \, dS = \int_0^(2\pi) \int_0^1 (1+u^2) \, du \, dv = 2\pi \int_0^1 (1+u^2) \, du = \boxed{\frac{8\pi}3}`