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I couldn't develop the parametric curve nor I didn't understand very well

This is the question:

1

There is the answer:

2

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?

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Goldman7911
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1 Answers1

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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.

Mark Fantini
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