In Q1, the boundaries becomes the following:

y = x becomes v = u,

y = 2 becomes v = \frac{2}{u+1}

x = 0 becomes either u = 0 or v = -1 but since v > 0 its the first choice.

These three curves in the (u,v) plane enclosed a closed region.

Q2. Using Greens theorem in the plane

\int P dx + Q dy = \int \int Qx - Py dA

If we choose Qx = y^2 and Py = -x^2 so that

P = - x^2 y and Q = x y^2 and parameterize the ellipse as

x = a \cos t and y = b \sin t then

\int P dx + Q dy = \int_0 ^{2\pi} (a^3 b + a b^3)\sin^2t \cos^2t dt from which the result follows.

Hope this helps