Feel free to consult the details in my answer to 24438105 if you wish to compute the line integral directly. I don't see any specification of which method to use, so I'll do it the faster way.
By Green's theorem,
[tex]\displaystyle \int_C (x^2+y^2)\,\mathrm dx + 3xy^2\,\mathrm dy = \iint_D \frac{\partial(3xy^2)}{\partial x} - \frac{\partial(x^2+y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \iint_D (3y^2-2y)\,\mathrm dx\,\mathrm dy[/tex]
Since D is a disk with radius 2 centered at the origin, consider converting to polar coordinates using x = r cos(t ) and y = r sin(t ). Then
[tex]D = \left\{(r,\theta) \mid 0\le r\le 2\text{ and }0\le t\le2\pi\right\}[/tex]
[tex]x^2+y^2 = r^2[/tex]
[tex]\mathrm dx\,\mathrm dy = r\,\mathrm dr\,\mathrm dt[/tex]
[tex]\implies \displaystyle \iint_D (3y^2-2y) \,\mathrm dx\,\mathrm dy = \int_0^{2\pi}\int_0^2 (3r^2\sin^2(t)-2r\sin(t))r\,\mathrm dr\,\mathrm dt \\\\ = \int_0^{2\pi} \int_0^2 (3r^3\sin^2(t)-2r^2\sin(t))\,\mathrm dr\,\mathrm dt \\\\ = \int_0^{2\pi} \left(12\sin^2(t)-\frac{16}3\sin(t)\right)\,\mathrm dt = \boxed{12\pi}[/tex]
