Question

Given the following vector field and oriented curve​ C, evaluate ModifyingBelow Integral from nothing to nothing With Upper C Bold Upper F times Upper T font size decreased by 5 ds. Bold Upper F equals left angle x comma y right angle on the parabola Bold r (t )equals left angle 6 t comma 11 t squared right angle​, for 0 less than or equals t less than or equals 1 The value of the line integral of F over C is nothing. ​(Type an exact​ answer, using radicals as​ needed.)

Answer 1

The value of the line integral is
`(157)/(2)`.

Explanation:

Given a path C, with parametrization r(t) for
`t_0\leq t \leq t_1`, we have that

`\int_(C)F dr = \int_(t_0)^(t_1) F(r(t)) \cdot r'(t) dt` where
`\cdot` is the dot product between two vectors.

In our case, we have
`r(t) = (6t,11t^2), 0\leq t\leq 1`. Then
`r'(t) = (6,22t)`. In this case we have that F(x,y) = (x,y). Then,
`F(r(t)) = (6t, 11t^2)`

So

`\int_(C)F dr = \int_(0)^(1) (6t,11t^2)\cdot(6,22t) dt = \int_(0)^(1) 36t+11\cdot 22 t^3= \left.(36(t^2)/(2)+11\cdot 22 (t^4)/(4))\right|_(0)^(1) = (36)/(2)+(11\cdot 22)/(4)= (157)/(2)`