Let $T$ be the portion of the surface $x^2= y^2 + z^2$ lying between the planes $x= 0$ and $x= 2$ and above the plane $z=0$. Calculate the surface integral
$$\iint_T (2 + x^2 y^2)\ \mathrm{d}S $$ i.e mass of surface $T$ if its density is $2 + x^2y^2$.
I don't know which formula I should use in this case. Thanks all!
You could find a parametrization of the cone, such as \begin{align} x &= r \\ y &= r\cos{t} \\ z &= r\sin{t} \end{align} In other words, you have found a function $\varphi(r,\,t)$ wich is: $$ \varphi(r,\,t) = (r,\, r\cos(t),\, r\sin(t)) $$
Then, you can solve the first type surface integral using: $$ \renewcommand{\d}{\,\mathrm{d}} $$ $$ \iint_M f(x,\,y,\,z)\d S = \iint_U f\left(\varphi\left(r,\,t\right)\right)\cdot\|\varphi_r\times\varphi_t\|\d (r,\,t)$$
Where $U$ are the boundaries of your new parametrization, in other words, you have:
\begin{align} \varphi_r &= (1,\, \cos(t),\, \sin(t)) \\ \varphi_u &= (0, -r\sin(t),\, r\cos(t)) \\ \|\varphi_r\times\varphi_t\| &= \sqrt{2}r \end{align}
And the field $f$ evaluated in $\varphi$ is $$ f(\varphi(r,\,t)) = 2 + r^4\cos^2(t) $$
Now, the last thing you need are the new boundaries, wich are: \begin{align} r &\in [0,\,2] \\ t &\in [0, \pi] \end{align}
So in the end, you only need to find the answer for: $$ \sqrt{2} \int_0^\pi\int_0^2 2r + r^5\cos^2(t) \d r\d t $$