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?