How do I calculate an integral of $F$ where $F = \langle y,x^2\rangle$?

Question

I have to calculate the above integral and I'm not sure how to do it. First I find the curl of $\langle y,x^2\rangle$ and I get:

$$\langle0,0,2x-1\rangle$$

Therefore it's not conservative because it's not all $0$s and I can't solve the integral.

Is that right? I feel like that isn't right.

Answer 1

$y = 2 + \frac{x^2}{2} \,$ (from $(0,2)$ to $(2,4))$

Parametrize it using $x = t$, $y = \frac{t^2}{2} + 2$

So your points on the curve are given by $(t, \frac{t^2}{2} + 2)$. Starting point is at $t = 0$ and the end point is at $t = 2$.

Now vector field is $(y, x^2)$ or $ (\frac{t^2}{2} + 2, t^2)$.

$r'(t) = (1, t)$.

Now you can do a dot product for your line integral from $t = 0$ to $t = 2$

Answer 2

You can parametrize the path using $r(x)=(x,2+\frac{x^2}{2}),\,x\in[0,2]$. Hence, $r'(x)=(1,x),\,F(r(x))=F(x,2+\frac{x^2}{2})=(2+\frac{x^2}{2},x^2)$, and then proceed by definition of the line integral.