I couldn't develop the parametric curve nor I didn't understand very well
This is the question:
There is the answer:
I tried $\int_{0}^{1}(3y-\sqrt{z})ds = \int_{0}^{1}(2y)ds$ because $z=y^2$ but didn't work
I used first this parametric:
A(1,0,0) to B (1,2,4)
x(t) = 1
y(t) = 2t
z(t) = 4t
I tried more things but didn't find the correct answer. Someone please could help me understand how to find the correct parametric curve to apply in integral?
He has given you the parametrization, namely $x=1$ and $z=y^2$. This parametrization is in terms of $y$. This means you substitute whatever is inside the integral with these given parameters and compute $ds$, which is the length of the tangent vector: $$\| {\mathbf r}'(y) \| = \| (1,y,y^2) \| = \sqrt{1+(2y)^2} = \sqrt{1+4y^2}.$$ Notice that $0 \leq y \leq 2$. Write the integral in terms of $y$ and you will find the desired answer.