use the Green's Theorem to calculate F dr, where C is the boundary of the region enclosed by y=x^2 and y=cube root of x, and F(x,y)=<x^2-y^3,x^3-y^2>. Please help me.
$\displaystyle F(x,y) = \langle P,Q\rangle = \langle x^2-y^3,x^3-y^2\rangle$
$\displaystyle \frac{\partial Q}{\partial x} = 3x^2$
$\displaystyle \frac{\partial P}{\partial y} = -3y^2$
So, $\displaystyle \int\int_A\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA = 3\int\int_A(x^2+y^2)\,dA$
Keeping in mind that $\displaystyle \sqrt[3]{x}\geq x^2$ on $\displaystyle [0,1]$, the bounds are:
$\displaystyle x=0...1$
$\displaystyle y=x^2...\sqrt[3]{x}$
So you need to integrate:
$\displaystyle 3\int_0^1\int_{x^2}^{\sqrt[3]{x}}(x^2+y^2)\,dy\,dx$