<s>[INST] evaluate the surface integral $f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = \sqrt{x^2 + y^2}$ with $z \leq 2$ Evaluate the surface integral of scalar function $\int_{S} f dS$
$f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = \sqrt{x^2 + y^2}$ with $z \leq 2$
my attempt
$f(x,y,z) = x^2 + y^2$ and $z = \sqrt{x^2 + y^2}, z \leq 2$
$\frac{dz}{dx} = \frac{x}{\sqrt{x^2 + y^2}}$ and $\frac{dz}{dy} = \frac{y}{\sqrt{x^2 + y^2}}$
$ds = \sqrt{\left(\frac{dz}{dx} \right)^2 + \left(\frac{dz}{dy} \right)^2} dx dy= \sqrt{\frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2}}dx dy=dx dy$
$\int_S f dS = \iint(x^2 + y^2)dx dy$, such that $x^2 + y^2 = 4, z \leq 2$
$\int_{0}^{2\pi} \int_{0}^{2} r^2 r dr d \theta = 8 \pi$
would this be right?
 [/INST] Your approach is almost correct but for one erroneous formula:
$\mathrm{d}S = \sqrt{z^2_x+z^2_y + 1} \ \mathrm{d}A $
$$
\begin{align*}
\mathrm{d}\vec{S} = \langle -z_x,-z_y, \ 1 \rangle \ \mathrm{d}A
\end{align*}
$$
therefore, $$|\mathrm{d}\vec{S}| =  \sqrt{z^2_x+z^2_y + 1} \ \mathrm{d}A $$
$z_x$ and $z_y$ values what you got are correct.
Let's proceed:
$$ 
\begin{align*}
\iint_S f\mathrm{d}S &= \iint_S x^2+y^2 \ \sqrt{z^2_x+z^2_y + 1} \ \mathrm{d}A \\
\\
&= \iint_S x^2+y^2 \sqrt{\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}+1} \ \mathrm{d}x\mathrm{d}y \\
\\
\end{align*}
$$
Now switch to polar coordinates:
$$
\begin{align*}
\iint_S f\mathrm{d}S &= \iint_S r^2\sqrt{2} \cdot r\mathrm{d}r\mathrm{d}r\theta \\
\\
&= \sqrt{2}\int_{0}^{2\pi}\int_{0}^{2}r^3 \ \mathrm{d}r\mathrm{d}\theta \\
\\
&= 8\sqrt{2}\pi
\end{align*}
$$
</s>