Answer
$8$
Work Step by Step
Given:
$r(t) = (t^2)i + (t^3)j + (-2t)k$
we get:
$x = t^2$
$y = t^3$
$z = -2t$
Derive $r(t)$ to get r'(t):
$r'(t) = (2t, 3t^2, -2)$
To calculate the line integral of $f$ we use:
$\int_c F \times dr = \int_c F(r(t)) \cdot r'(t) dt$
$ = \int_0^2 (t^2 + (t^3)^2, (t^2)(-2t), t^3 - 2t)\cdot(2t, 3t^2, -2) dt $
$ = \int_0^2 (t^2 + t^6, -2t^3, t^3-2t) \cdot (2t, 3t^2, -2) dt $
$ = \int_0^2 (2t^3 + 2t^7 - 6t^5 - 2t^3 + 4t) dt $
$ = \int_0^2 (2t^7 - 6t^5 + 4t) dt$
$ = (\frac{2}{8}t^8 - t^6 + 2t^2)_0^2$
$ = (\frac{2}{8}(2)^8 - (2)^6 + 2(2)^2) - (\frac{2}{8}(0)^8 - (0)^6 + 2(0)^2)$
$ = 8$
