By Green's theorem,
[tex]\displaystyle\int_{x^2+y^2=16}(3y-e^{\sin x})\,\mathrm dx+(7x+y^4+1)\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_{x^2+y^2\le16}\frac{\partial(7x+y^4+1)}{\partial x}-\frac{\partial(3y-e^{\sin x})}{\partial y}\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle4\iint_{x^2+y^2\le16}\mathrm dx\,\mathrm dy[/tex]
The remaining integral is just the area of the circle; its radius is 4, so it has an area of 16π, and the value of the integral is 64π.
We'll verify this by actually computing the integral. Convert to polar coordinates, setting
[tex]\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta[/tex]
The interior of the circle is the set
[tex]\{(r,\theta)\mid0\le r\le4\land0\le\theta\le2\pi\}[/tex]
So we have
[tex]\displaystyle4\iint_{x^2+y^2\le16}\mathrm dx\,\mathrm dy=4\int_0^{2\pi}\int_0^4r\,\mathrm dr\,\mathrm d\theta=8\pi\int_0^4r\,\mathrm dr=64\pi[/tex]
as expected.
