line integral over a triangle

[mathstresser]

Green’s Theorem is
\(\displaystyle \L\\\int_{c}P dx + Qdy = \int_{D} ((dQ)/(dx)-(dP/dy)) dA\)

Evaluate the line integral by two methods: (a) directly and (b) using Green’s Theorem.

\(\displaystyle \L\\\int_{C}xydx + x^2y^3dy\)
C is a triangle with vertices (0,0), (1,0) and (1,2)

\(\displaystyle \L\\ P=xy, Q=x^2y3\)

When I think, "evaluate the integral directly," this is what I get:

\(\displaystyle \L\\\1/2x^2y + 1/4x^2y^4\)

(Is that right?) But I don’t know how to evaluate it.

Then for Green’s Theorem, I have

dQ/dx=2x, dP/dy=x

\(\displaystyle \L\\\int_{d} (dQ/dx)minus(dP/dy) dA= \int_{d} 2x-x dA= \int_{D} x dA\)

This next step I’m not sure about, but this is what I get: (The dA is the part I really don’t know about, but the other could be wrong too…)

\(\displaystyle \L\\\int_{0,0}^{1,0} xdA + \int_{1,0}^{1,2} xdA+ \int_{1,2}^{0,0} xdA\)

I know this is long, and I’m sorry. But these steps are driving me crazy! This is everything I have.

 

[galactus]

For the Green's Theorem part:

\(\displaystyle \L\\f(x,y)=xy\)

\(\displaystyle \L\\g(x,y)=x^{2}y^{3}\)

\(\displaystyle \L\\\int\int_{C}\left[\frac{\partial{g}}{\partial{x}}-\frac{\partial{f}}{\partial{y}}\right]\)

\(\displaystyle \H\\\int_{0}^{1}\int_{0}^{2x}(2xy^{3}-x)dydx\)