Example of using Green's Theorem
Evaluate the line integral `int_C 1/2 y^6dx + (6/5 x^5+3xy^5)dy` along the curve `C` from (0,0) along `y=(x+2)^2 -4` to (1,5) and then using `y=5x` back to (0,0).

We might note that we could just evaluate the line integral via a program such as the topics15s4demo.pl program, here, though not shown.

This has the form of Green's Theorem, namely, it looks like `int_C Mdx+Ndy`
    where `M(x,y)=1/2 y^6` and `N(x,y)= 6/5 x^5+3xy^5`.

Therefore, by Green's Theorem, we can rewrite the line integral as the double integral over the surface in the form
`int_Rint (delN)/(delx)-(delM)/(dely) dA`. However, we see that

`(delN)/(delx) = 6x^4+3y^5` and

`(delM)/(dely) = 3y^5`. So

`int_Rint (delN)/(delx)-(delM)/(dely) dR = int_Rint 6x^4+3y^5 - 3y^5 dA`.

This simplifies to

`int_Rint 6x^4 dA`.

Fortunately, our region R is "nice" in that we can describe it by saying that x goes between 0 and 1, the lower bound of the region is a function of x, namely,

`y=(x+2)^2-4 = x^2 +4x +4 - 4 = x^2+4x`

and the upper bound of the region is a function of x, namely,

`y=5x`.

Therefore, we can convert the double integral to an iterated integral as in

`int_Rint 6x^4 dA = int_0^1 int_(x^2+4x)^(5x)6x^4 dy dx`.

Dealing with the inside integral, the anti-derivative with respect to `y` is

`6x^4y` and we need to find

`(6x^4y) |_(x^2+4x)^(5x) = (6x^4)(5x) - (6x^4)(x^2+4x)`. But that simplifies to

`30x^5 -6x^6-24x^5 = 6x^5-6x^5`. Thus, our problem now becomes

`int_0^1 6x^5-6x^6 dx`. However, the anti-derivative of `6x^5-6x^6` is

`x^6 - 6/7x^7` which we need to evaluate between 0 and 1 as in

`x^6 - 6/7x^7)|_0^1 = (1 - 6/7) - (0-0) = 1/7`.

Thus, for the give closed curve C we have

`int_C 1/2 y^6dx + (6/5 x^5+3xy^5)dy = int_Rint 6x^4+3y^5 - 3y^5 dA = int_Rint 6x^4 dA = int_0^1 int_(x^2+4x)^(5x)6x^4 dy dx = 1/7`.