Let $S$ be the bounded surface of the cylinder $x^2+y^2=1$ cut by the planes $z=0$ and $z=1+x$ Then how to show that the value of the surface integral $∬3z^2 d \sigma $ over $S$ is equal to $\int_{0}^{2\pi}(1+cos \theta)^3d\theta$
I tried using $\int\int_{S}g d\sigma =\int\int_{R}g \frac{|\nabla f|}{|\nabla f . p|}dA$ but I dont know what is what.
The body on which you want to carry on the surface integral can be parametrized as
$$\;r(t)=(\cos\theta,\,\sin\theta,\,t)\;,\;\;0\le \theta\le2\pi\;,\;\;0\le t\le 1+\cos\theta$$
and we thus get:
$$r_t=(0,0,1)\;,\;\;r_\theta=(-\sin\theta,\,\cos\theta,\,0)\implies$$
$$\implies r_t\times r_\theta=\begin{vmatrix}i&j&k\\0&0&1\\-\sin\theta&\cos\theta&0\end{vmatrix}=(-\cos\theta,\,-\sin\theta,\,0)\implies ||r_t\times r_\theta||=1$$
and our integral becomes
$$\int_0^1\int_0^{2\pi}\int_0^{1+\cos\theta} 3z^2dzd\theta dr=\int_0^{2\pi}(1+\cos\theta)^3d\theta =$$