TITLE: Line Integral parametrised curve, unparametrised function
QUESTION [0 upvotes]: Integrate f over the given curve.
$$ f(x,y,z) = x \sqrt{y} - 3z^2 $$
$$r(t) = \cos(2t)i+\sin(2t)j+5tk$$
lower and upper bounds of t are [0, 2$\pi$], respectively.
I took the derivative of r(t)
$$ r'(t) = -2\sin(t)\,i+2\cos(2t)\,j+5k $$
Then I substituted the $x,y$ and z values into $f(x,y,z)$ because I've seen this equation before:
$$ \int_{0}^{2\pi} F(r(t)) ||r'(t)|| dt$$
$$ \int_{0}^{2\pi}( \cos 2t \sqrt{\sin 2t}-3(5t)^2 )\sqrt{(-2\sin t)^2 + (2\cos 2t )^2+5^2} \,dt $$
I haven't simplified it yet. I can't help but feel like I've made a mistake though because that integral looks nasty! I was hoping someone could check my work but also answer my question (below).
Are these equivalent statements? 
$$ f(x,y,z) = x \sqrt{y} - 3z^2 $$
$$ f(x,y,z) = x \sqrt{y}\,i - 3z^2\,j + 0k$$

REPLY [0 votes]: $f(x,y,z) = x \sqrt(y) - 3z^2$ is not the same as $f(x,y,z) = x \sqrt{y} \mathbf i - 3z^2\mathbf j + 0\mathbf k$
One is a scalar and the other is a vector.
$r'(t) = -2\sin(2t) \mathbf i + 2\cos(2t)\mathbf j +5\mathbf k\\
||r'(t)|| = \sqrt{(-2\sin(2t))^2 + (2\cos(2t))^2 +(5)^2} = \sqrt{41}$
$\int (\cos(2t)\sqrt{\sin(2t)} - 75t^2)\sqrt{41} \, dt$
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