Assuming the curves are traced in a counterclockwise direction we can simply apply Green's Theorem directly.

1) ∫c (ln(x+y) dx - x² dy) = ∫∫E (-2x - (1 / (x+y))) dA

where E is the region enclosed by c. The limits for that region are [1,3] for x and [1,4] for y, as evidence by the vertices of the retangular region. So

∫[1,4] ∫[1,3] (-2x - (1 / (x+y))) dx dy

That's easy enough to calculate.

Similarly

2) ∫c (x²y dx + 0 dy) = ∫∫E -x² dA

You could integrate in Cartesian coordinates, but I think polar is easier. So converting we have

-∫[0,1] ∫[0,2π] r³ cos²(θ) dθ dr

where the extra r came from the Jacobian of the transformation. Again, that's easy to calculate (especially if you know the trick that ∫[a,b] ∫[c,d] f(x)g(y) dx dy = (∫[a,b] f(x) dx)(∫[c,d] g(y) dy))

First you opt to remodel the curve to parametric type: x^2+y^2=9 => x=3cost, y=3sint, dx=3sint, dy=-3cost Now you've the ?y^3dy = ?(3sint)^3*3sint dt This imperative evaluates to three^4*a million/32*[12x - 8sin2x + sin4x] and think ofyou've got were given -?x^3dx = ?(3cost)^3*-3cost dt This evaluates to -3^4*a million/32*[12x + 8sin2x + sin4x] So your effect is -3^4*a million/32*[16sin2x + 2sin4x] desire that facilitates!