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 16Ï€, and the value of the integral is 64Ï€.
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.
