r(t) = <cos(t), 2t, sin(t)> from 0 to 2pi (included).
Compute the integral from 0 to 2pi of r(t).
The Attempt at a Solution
We haven't learned this in class yet, so I am getting a head. My idea is to simply put r(t) in and integrate term by term so that the final answer is
sin(t), t^2, -cos(t) and evaluate it from 0 to 2pi which gets
<0, 4pi^2, -1> - < 0, 0, -1>
which results in
<0, 4pi^2, 0>
And then I take the magnitude? Eh, answer should not be a vector I assume since this is the area of the space.