I have to evaluate $I$ $$ I=\iint\limits_S (x^2+y^2)\ dS\ \ \text{where $S$ is the surface of the cone}\ \ \sqrt{x^2+y^2}\leqslant z\leqslant 1 $$
My attempt:
$$
\begin{aligned}
&I=\iint\limits_{S_1} (x^2+y^2)\ dS_1 + \iint\limits_{S_2} (x^2+y^2)\ dS_2,\ \ \text{where}\\
&S_1 \text{ is the side surface of the cone},\ \ S_2 \text{ is the base surface of the cone}
\end{aligned}
$$
The first integral is straigtforward:
$$
\begin{aligned}
&I_1=\iint\limits_{S_1} (x^2+y^2)\ dS_1=\iint\limits_{M_1}(x^2+y^2)\sqrt{(z'_x)^2+(z'_y)^2+1}\ dxdy,\ \ \text{where}\ z=\sqrt{x^2+y^2}\\
&I_1=\sqrt{2}\iint\limits_{M_1}(x^2+y^2)\ dxdy=\sqrt{2}\int\limits_0^{2\pi}d\phi\int\limits_0^1r^2\cdot r\ dr=\frac{\pi\sqrt{2}}{2}
\end{aligned}
$$
However, the answer I got for the second integral and the problem book answer don't match.
$$
\begin{aligned}
&\text{Since $z=1$, we have}\ \ I_2=\iint\limits_{S_2} (x^2+y^2)\ dS_2=\int\limits_\sigma(x^2+y^2)\ dl,\ \ \text{where}\ \ \sigma\ \ \text{is a circle}\ \ x^2+y^2=1\\
&I_2=\int\limits_\sigma(x^2+y^2)\sqrt{(x'_\phi)^2+(y'_\phi)^2}\ d\phi=\int\limits_0^{2\pi}(\cos^2\phi+\sin^2\phi)\cdot 1\ d\phi=2\pi
\end{aligned}
$$
So, my final answer is $I=I_1+I_2=\pi\left(\frac{\sqrt{2}}{2}+2\right)$.
However, the correct answer is $I=\pi\left(\frac{\sqrt{2}}{2}+\frac{1}{2}\right)$.
That means, I must have made a mistake while evaluating $I_2$, but I have no idea what is wrong. I would greatly appreciate it if someone pointed it out to me.
Note the second integral is over the unit circle, not along the circumference,
$$I_2=\int\limits_{x^2+y^2<1}(x^2+y^2)dxdy =\int_0^{2\pi}\int _0^1 r^2rdrd\phi=\frac\pi2 $$