Parameterize [tex]C[/tex]
[tex]x=\cos t[/tex]
[tex]y=\sin t[/tex]
with [tex]0\le t\le2\pi[/tex]. Then the line integral of [tex]\langle2x^3-y^3,4x^3+y^3\rangle[/tex] along [tex]C[/tex] is
[tex]\displaystyle\int_0^{2\pi}\langle2\cos^3t-\sin^3t,4\cos^3t+\sin^3t\rangle\cdot\langle-\sin t,\cos t\rangle\,\mathrm dt[/tex]
[tex]\displaystyle=\int_0^{2\pi}(4\cos^4t-2\cos^3t\sin t+\cos t\sin t^3+\sin^4t)\,\mathrm dt=\boxed{\frac{15\pi}4}[/tex]
By Green's theorem, the line integral is equivalent to
[tex]\displaystyle\iint_{x^2+y^2\le1}\left(\frac{\partial(4x^3+y^3)}{\partial x}-\frac{\partial(2x^3-y^3)}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_{x^2+y^2\le1}(12x^2+3y^2)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^1(12r^2\cos^2\theta+3r^2\sin^2\theta)r\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_0^1(9r^3\cos^2\theta+3r^3)\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{15\pi}4}[/tex]
