Solution
\begin{align*}
\mathbf{r}_u &= \langle \cos v, \sin v, 0 \rangle \\
\mathbf{r}_v &= \langle -u\sin v, u\cos v, 1 \rangle \\
\left|\mathbf{r}_u\times\mathbf{r}_v\right| &= \left|\langle \sin v, -\cos v, u\rangle\right| = \sqrt{1+u^2} \\ \\
A(S) &= \int_{-\pi}^\pi \int_0^3 \sqrt{1+u^2}\,du\,dv = \int_{-\pi}^\pi\,dv\int_0^3\sqrt{1+u^2}\,du \\
&= 2\pi\bigg[ \frac{1}{2}u\sqrt{1+u^2}+\frac{1}{2}\ln\left|u+\sqrt{1+u^2}\right| \bigg]_0^3 = \color{answer}\mathbf{\pi(3\sqrt{10}+\ln(3+\sqrt{10}))}
\end{align*}
\begin{align*}
z &= 4-x-2y \\ \\
A(S) &= \int_0^2\int_0^{4-2y}\sqrt{1+\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2}\,dx\,dy
= \int_0^2\int_0^{4-2y}\sqrt{1+(-1)^2+(-2)^2}\,dx\,dy \\
&= \int_0^2\int_0^{4-2y}\sqrt{6}\,dx\,dy = \sqrt{6}\int_0^2(4-2y)\,dy \\
&= \sqrt{6}\bigg[4y-y^2\bigg]_0^2 = \color{answer}\mathbf{4\sqrt{6}}
\end{align*}
Use cylindrical coordinates to parametrize the surface.
\begin{align*}
\mathbf{r}(\theta,z) &= \langle 2\cos\theta,2\sin\theta,z\rangle,\quad 0\le \theta\le 2\pi,~-1\le z\le 3 \\
\mathbf{r}_\theta &= \langle -2\sin\theta,2\cos\theta,0\rangle \\
\mathbf{r}_z &= \langle 0,0,1 \rangle \\
\left|\mathbf{r}_\theta\times\mathbf{r}_z\right| &= \left|\langle 2\cos\theta,2\sin\theta,0\rangle\right| = \sqrt{4\cos^2\theta+4\sin^2\theta} = 2 \\ \\
A(S) &= \int_0^{2\pi}\int_{-1}^3 2\,dz\,d\theta = 2\cdot 2\pi\cdot 4 = \color{answer}\mathbf{16\pi}
\end{align*}
\begin{align*}
A(S) &= \iint_D\sqrt{1+\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2}\,dA \\
&= \iint_D \sqrt{1+\tfrac{1}{4}(x^2+y^2)}\,dA = \int_0^{2\pi}\int_0^2 \sqrt{1+\tfrac{1}{4}r^2}r\,dr\,d\theta \\
&= 2\pi \int_0^2\sqrt{1+\tfrac{1}{4}r^2}r\,dr = 2\pi\bigg[ \frac{4}{3}(r^2+1)^{3/2} \bigg]_0^2 = \color{answer}\mathbf{\frac{8\pi}{3}(2\sqrt{2}-1)}
\end{align*}