Evaluating Line Integrals with Green's Theorem

[gtfitzpatrick]

Homework Statement

Let C be the boundary of the region bounded by the curves y=$x^{2}$ and y=x. Assuming C is oriented counter clockwise, Use green's theorem to evaluate the following line integrals (a) $\oint(6xy-y^2)dx$ and (b) $\oint(6xy-y^2)dy$

Homework Equations

The Attempt at a Solution

$\int^{0}_{1} 6x^2 - x^2$
$\int^{0}_{1} 5x^2$ = -$\frac{5}{3}$
and
$\int^{1}_{0} 6x^3 - x^4$ = $\frac{6}{4} - \frac{1}{5} = \frac{13}{10}$

so $\oint$ = -$\frac{11}{30}$

but
$\int\int_{R}$ $\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}$)dxdy
M=6xy-$y^{2}$ and N=0
$\frac{\partial M}{\partial y} 6x-2y$

$\int\int_{R}$(6x-2y)dxdy

$\int^{1}_{0} [ \int^{x}_{y=x^2} (6x-2y)dy] dx$

$\int^{1}_{0} 5x^2 - 6x^3 - x^4 dx$

= $\frac{-1}{30}$
anyone got any idea what I am doing wrong here!stumped

 

[LCKurtz]

Homework Statement

Let C be the boundary of the region bounded by the curves y=$x^{2}$ and y=x. Assuming C is oriented counter clockwise, Use green's theorem to evaluate the following line integrals (a) $\oint(6xy-y^2)dx$ and (b) $\oint(6xy-y^2)dy$

Homework Equations

The Attempt at a Solution

$\int^{0}_{1} 6x^2 - x^2$
$\int^{0}_{1} 5x^2$ = -$\frac{5}{3}$
and
$\int^{1}_{0} 6x^3 - x^4$ = $\frac{6}{4} - \frac{1}{5} = \frac{13}{10}$

so $\oint$ = -$\frac{11}{30}$

Poorly written, but assuming you are doing the circuit integral -11/30 is correct.

but
$\int\int_{R}$ $\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}$)dxdy
M=6xy-$y^{2}$ and N=0
$\frac{\partial M}{\partial y} 6x-2y$

$\int\int_{R}$(6x-2y)dxdy

$\int^{1}_{0} [ \int^{x}_{y=x^2} (6x-2y)dy] dx$

$\int^{1}_{0} 5x^2 - 6x^3 - x^4 dx$

= $\frac{-1}{30}$
anyone got any idea what I am doing wrong here!stumped

Check your sign on that $x^4$ term in the second to last line. And don't you want $-M_y$ for Green's theorem?