M is the surface $z=x^2+y^2$ with standard orientation for $x^2+y^2\leq 1$ and $\varphi = 4x^2ydy+z^2dz$
I'd like to verify that $\int_Md\varphi=\int_{\partial M}\varphi$, which I did, but according to the book the solution is $\pi$ while I got $0$ computing both cases.
$(1)$ Here I calculate $\int_M d\varphi$.
I take the change of coordinates $k(r,\theta)=(r\cos \theta,r\sin\theta,r^2)$, then $k_r=(\cos\theta,\sin\theta,2r)$ and $k_\theta=(-r\sin\theta,r\cos\theta,0)$. Since $d\varphi= 8xydxdy$ I have $$d\varphi(k_r,k_y)=8(r\cos\theta)(r\sin\theta)(r\cos^2 \theta+r\sin^2\theta)\\=8r^3\sin\theta\cos\theta.$$
Then $$\int_M d\varphi=\int_0^{2\pi}\int_0^1 8r^3 \sin \theta \cos \theta dr d\theta\\=2\int_0^{2\pi}\sin\theta\cos\theta(r^4|_0^1)d\theta=2\int_0^{2\pi}\sin\theta\cos\theta d\theta = \sin^2\theta|_0^{2\pi}=0$$
$(2)$ Now if I do it calculating $\int_{\partial M}\varphi$ ...
I fix $r=1$ and this gives the parameterization $r(\theta)=(\cos\theta,\sin\theta,1).$ Then $$\int_{\partial M} \varphi = \int_0^{2\pi}4\cos^2\theta\sin\theta(\cos\theta)d\theta\\=4\int_0^{2\pi}\sin\theta\cos^3\theta d\theta=\color{red}{-}cos^4\theta|_0^{2\pi}=0.$$
What did I do wrong?