1. ## line integral

I want to compute the line integral

integral (2ydy-ydx) where C is the boundary of the half disc (x^2)+(y^2) less than or equal to 1 with y greater than or equal to 0 traversed in the positive sense by parametrising the boundary curve (there are two pieces, a straight line segment and a semi-circle) and then evaluating the integral directly.

I did this:
integral (2ydy-ydx)
I use the following:
integral (f1(x,y)dx+f2(x,y)dy)= double integral (df2/dx-df1/dy)dA

Thus, I have I=double integral(d/dx(2y)+d/dy(y))dA
I=double integral (1)dA=integral (from 0 to pi)dtheta integral (from 0 to 1) 1dr=integral (from 0 to pi) 1dtheta=pi.

I got an answer of pi for this question, but it seems too easy. Is this correct?

I used Green's Theorem by the way.

2. You need to find,
$\oint_C -ydx+2ydy$
Where $C$ is a positively oriented, piecewise smooth, closed simply connected region.
Thus, we can use Green's theorem.
$\oint_C -ydx+2ydy=\int _D\int \left( \frac{\partial (2y)}{\partial x}-\frac{\partial (-y)}{\partial y}\right) dA=\int_D\int dA$
Thus it is simply the area of the region.
In this case,
$\frac{1}{2}\pi$