Treat [tex]\mathcal C[/tex] as the boundary of the region [tex]\mathcal S[/tex], where [tex]\mathcal S[/tex] is the part of the surface [tex]z=2xy[/tex] bounded by [tex]\mathcal C[/tex]. We write
[tex]\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r[/tex]
with [tex]\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)[/tex].
By Stoke's theorem, the line integral is equivalent to the surface integral over [tex]\mathcal S[/tex] of the curl of [tex]\mathbf f[/tex]. We have
[tex]\nabla\times\mathbf f=(-2z,-3x^2,-1)[/tex]
so the line integral is equivalent to
[tex]\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S[/tex]
[tex]=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv[/tex]
where [tex]\mathbf s(u,v)[/tex] is a vector-valued function that parameterizes [tex]\mathcal S[/tex]. In this case, we can take
[tex]\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex]. Then
[tex]\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv[/tex]
and the integral becomes
[tex]\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi[/tex]
