Line integral counterclockwise.

Question

let $F$ be a vector field, such that $$F(x,y)=\langle x^2y^2,xy^2\rangle$$

Calculate the integral of $F$ along the closed formed by the line $x=1$ and the parabola $y^2=x$.

first, we have to parametrize the line and the parabola, I will choose the curve $C_1$ for the line and $C_2$ for the parabola

$$\cases{C_1(t)=\langle 1,t\rangle \\ C_2(t)=\langle t^2,t\rangle} \text{ and } \cases{F(C_1(t))=\langle t^2,t^2\rangle \\ F(C_2(t))=\langle t^6,t^4 \rangle}$$ So we get $$\cases{F(C_1(t)) \cdot C_1'(t)=t^2 \\ F(C_2(t))\cdot C_2'(t)=2t^7+t^4}$$

and remember that $$\int_C F=\int_{C_1}F+\int_{C_2}F$$ but now I have no idea how to find the bounds, and even if I did, I don't know how can I calculate this integral counterclockwise.

Answer 1

Draw it out, pick a point, and move along the boundary of the enclosed region counter clockwise. For example if we start at the point $p = (1,-1)$ then our integral over $C_1$ is

$$ \int_{-1}^{1} t^2 \text{d}t =\frac{2}{3}\text{ .}$$

Clearly this runs from $C_1(-1)=(1,-1)$ to $C_1(1)=(1,1).$

Going over the parabola, we start at $C_2(1) = (1,1)$ and move to $C_2(-1) = (1,-1)$ thus we can integrate

$$ \int_{1}^{-1} 2t^7 + t^4 \: \text{d}t =-\frac{2}{5}\text{ .}$$

Summing these two integrals completes the loop: $$\int_CF=\frac{2}{3} - \frac{2}{5}=\frac{4}{15}$$

Answer 2

As it is closed curve, you can apply Green's theorem too.

$F(x,y)= (P, Q) = (x^2y^2,xy^2)$

Region is defined by $R: y^2 \leq x \leq 1$

So, $\displaystyle \int_C \vec F \cdot dr = \iint_R (Q_x - P_y) \ dA$

$ \displaystyle = \iint_R (y^2 - 2x^2y) \ dA$

The second part of the integral will be zero as the region is symmetric about x-axis and $x^2y$ is an odd function wrt $y$. So integral simplifies to,

$ \displaystyle = \int_{-1}^{1} \int_{y^2}^1 y^2 \ dx \ dy = \frac{4}{15}$