ASAP help Test in 2days. GReens theorem.

**ottmar0** · May 9th 2009, 12:28 PM

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.

**TheEmptySet** · May 9th 2009, 12:34 PM

Originally Posted by ottmar0

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.

Greens theorem tells us that if C is any peicewise simple closed curve then

$\oint _C \vec F \cdot d\vec r=\iint_D \left(\frac{\partial }{\partial x}(x^3-y^2)-\frac{\partial }{\partial x}(x^2-y^3) \right)dA=\iint_D(3x^2+3y^2)dA$

Can you finish from here

**redsoxfan325** · May 9th 2009, 12:35 PM

Originally Posted by ottmar0

use the Green's Theorem to calculate $\int_C F\,dr$ , where $C$ is the boundary of the region enclosed by $y=x^2$ and $y=\sqrt[3]{x}$ , and $F(x,y)=\langle x^2-y^3,x^3-y^2\rangle$ . Please help me.

$F(x,y) = \langle P,Q\rangle = \langle x^2-y^3,x^3-y^2\rangle$

$\frac{\partial Q}{\partial x} = 3x^2$

$\frac{\partial P}{\partial y} = -3y^2$

So, $\int\int_A\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA = 3\int\int_A(x^2+y^2)\,dA$

Keeping in mind that $\sqrt[3]{x}\geq x^2$ on $[0,1]$ , the bounds are:

$x=0...1$
$y=x^2...\sqrt[3]{x}$

So you need to integrate:

$3\int_0^1\int_{x^2}^{\sqrt[3]{x}}(x^2+y^2)\,dy\,dx$

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