Let me try...
First we need to find,
And is a box.
Thus,
Thus,
Thus,
Thus,
Question 2]
You are given,
To determine if it is conservative for well-behaved vector fields the necessary and sufficient conditions is to check if
Thus, the we have,
Thus if is.
It turns out that there is an easier way in the special case for two dimension vector fields.
Necessary and suffient for well-behaves fields,
Called the "cross-partials test"
(If you taken Differencial Equations you will note it is the same test to determine if a differencial equation is exact or not).
In this case,
Thus, it is conservative.