Let a_x and a_y be unit vectors along x and y directions, respectively. A vector function is given by F = a_x y - a_y x The line integral of above function

${\int }_{C}F\cdot dl$ Along the curve C, which follows the parabola y = x2 as shown below is ______ (rounded off to 2 decimal places)

The correct answer is -3.

Step 1: Understand the Given Vector Function and Path Differential

We are given the vector field:

$F=y{a}_x-x{a}_y$

The differential displacement vector in 2D Cartesian coordinates is:

$dl=dx{a}_x+dy{a}_y$

The dot product $F·dl$ is calculated as:

$F·dl=(y{a}_x-x{a}_y)·(dx{a}_x+dy{a}_y)=ydx-xdy$

Step 2: Identify Curve Parameters and Limits from the Image

From the given diagram, the curve $C$ follows the parabola:

$y=x^2$

Taking the differential on both sides gives:

$dy=2xdx$

Looking at the path direction in the image, the curve $C$ starts from $x=-1$ (where $y=1$) and moves along the parabola towards $x=2$ (where $y=4$). Thus, the limits for $x$ are from $-1$ to $2$.

Step 3: Evaluate the Line Integral

Substitute $y=x^2$ and $dy=2xdx$ into the integrand:

$F·dl=x^{2}dx-x\left(2xdx\right)=x^{2}dx-2x^{2}dx=-x^{2}dx$

Now, integrate over the limits $x=-1$ to $x=2$:

${\int }_{C}F·dl={\int }_{-1}^2-x^{2}dx=-{\left[\frac{x^3}{3}\right]}_{-1}^2$

Evaluating the boundary values:

${\int }_{C}F·dl=-\frac{1}{3}[2^3-(-1{)}^3]=-\frac{1}{3}[8-(-1)]=-\frac{1}{3}[9]=-3$

Hence, the line integral of the function along curve C is -3.