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

Answer:

The value of the line integral is $\frac{157}{2}$.

Step-by-step 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

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