I'm supposed to do the following problems, first using line integrals and then Green's theorem....but I keep getting different answers. Can someone please help me see my error? Thanks.

1a) Let C be the positively oriented boundry of the first quadrant part of the unit disk. Use the definition of the line integral to find:

$\displaystyle \int (xy)dx + (x+y)dy$

My Work:

$\displaystyle x=cos(t), y=sin(t), 0\leq t \leq \frac{\pi}{2}$ b/c it is only in the first quadrant

$\displaystyle \vec{F}<x,y>=(cos(t)*sin(t))$

$\displaystyle \vec{G}<x,y>=(cos(t)+sin(t))$

$\displaystyle \int_0^\frac{\pi}{2}[(cos(t)*sin(t))(-sin(t))+(cos(t)+sin(t))(cos(t))]dt$

$\displaystyle =\int_0^\frac{\pi}{2}[-sin^2(t)*cos(t)+cos^2(t)+sin(t)*cos(t)]dt$

$\displaystyle =\int[(-(u)^2)+u)]du+\int_0^\frac{\pi}{2}[\frac{1}{2}+\frac{cos(2t)}{2}]dt$

$\displaystyle [-\frac{sin^3(t)}{3}+\frac{sin^2(t)}{2}+\frac{t}{2}+ \frac{sin(2t)}{4}]\right|_0^\frac{}\pi{2}$

$\displaystyle =[-\frac{1}{3}+\frac{1}{2}+\frac{\pi}{4}]$

$\displaystyle =\frac{\pi}{4}-\frac{1}{6}$

1b) Now using Green's Theorem:

$\displaystyle \frac{\partial G}{\partial x}=1$

$\displaystyle \frac{\partial F}{\partial y}=x$

$\displaystyle \int_0^\frac{\pi}{2} \int_1^1[1-r cos(\theta)]r*dr*d\theta$

$\displaystyle \int_0^\frac{\pi}{2} \int_1^1[r-r^2 cos(\theta)]dr*d\theta$

$\displaystyle \int_0^\frac{\pi}{2}[\frac{r^2}{2}-\frac{r^3}{3}cos(\theta)]\right|_0^1 d\theta$

$\displaystyle \int_0^\frac{\pi}{2}[\frac{1}{2}-\frac{1}{3}cos(\theta)]d\theta$

$\displaystyle =[\frac{1}{2}\theta-\frac{1}{3}sin(\theta)]\right|_0^\frac{\pi}{2}$

$\displaystyle =\frac{\pi}{4}-\frac{1}{3}$