Applying Green's Theorem to evaluate line integral

Question

"Apply Green's Theorem to evaluate the line integral of F around positively oriented boundary"

$$F(x,y)=x^2yi+xyj$$

C: The region bounded by y=$x^2$ and y=4x+5

Answer 1

Green't Theorem implies,

$$ \oint_{C} F \cdot d{\vec{r}} = \iint_D (y-x^2) \ dA$$

Now you just need to choose how you wish you parametrize the region $D$ which is the area between the two curves you mentioned.

Say we wish to integrate with $dydx$ then we need $f_1(x) \leq y \leq f_2(x)$ and $\alpha \leq x \leq \beta$. Well $y$ which is your height, is bounded by what two curves? For the bounds of $x$, you've already given the intervals endpoints $-1,5$, so what is $\alpha, \beta$.

Answer 2

Greens theorem:

$\oint F(x,y)\cdot dr = \iint \nabla \times F dy dx$

Technically this is Stokes theorem but Stokes theorem is a generalization of Greens.

Lets show that Green's theorem is true:

$\oint F(x,y)\cdot dr$

The contour is in two pieces. lets use

$(x,y) = (t, t^2)$ from $-1$ to $5$ for the first piece and $(x,y) = (t,5 + 4t) from $5$ to $-1$ for the second.

$dr = (1,2t), (1,4)$ respectively

$\int_{-1}^5 (t^2)(t^2)(1)+ (t)(t^2)(2t)\ dt +\int_{5}^{-1} (t^2)(5+4t)(1)+ (t)(5+4t)4\ dt\\ \int_{-1}^5 3t^4 -4t^3 - 21t^2 - 20 t\ dt\\ \frac {3}{5} t^5 -t^4 - 7t^3 - 10 t^2|_{-1}^5\\ \frac 35 (5^5) - 5^4 - (7)(5^3) - 10(5^2) +\frac 35 + 1 - 7 + 10\\ 129\frac 35$

and applying Green's $\iint \nabla \times F dy dx$

$\int_{-1}^{5}\int_{x^2}^{5+4x} \frac {\partial}{\partial x} xy - \frac{\partial}{\partial y}x^2y\ dy\ dx\\ \int_{-1}^{5}\int_{x^2}^{5+4x} y-x^2\ dy\ dx\\ \int_{-1}^{5} \frac 12 y^2- yx^2\ |_{x^2}^{5+4x}\ dx\\ \int_{-1}^{5} \frac 12 (25 + 40x + 16x^2 - x^4)- (5+4x - x^2)x^2\ dx\\ \int_{-1}^{5} \frac 12 x^4 - 4x^3 +3 x^2 + 20 x + \frac {25}{2}\ dx\\ \frac 1{10} x^5 - x^4 + x^3 + 10 x^2 + \frac {25}{2}x |_{-1}^5\\ \frac 12 (5^4) - 5^4 + \frac 75 (125) + \frac 1{10} + 1 + 1 - 10 + \frac {25}{2}\\ 129\frac{3}{5}$

Hey it works!