Find the area of the cylinder given by $x^2+(y-1)^2=1$ between the xy-plane and the cone $\sqrt{x^2+y^2}+z=0$.
So parameterized the curve that describes the cylinder in the xy-plane:
$r(t)=(cost)i+(sint+1)j$
and then the height of surface must be $z=\sqrt{x^2+y^2}=\sqrt{2+2sint}$
I then solve the integral $ \int_{0}^{2\pi} \sqrt{2+2sint} \,dt$
But this does not give me the correct answer, which should be 16.
Edit (solution from my book that I don't understand):
The area asked for can be found by $$\begin{equation} \iint_{S_1}\,dxdy = 2\int_{C}1\,dS \end{equation}=2\int_{C}z\,ds$$ where $S_1$ is the surface of the cylinder between the xy-plane and the cone. $C$ is the curve of the cylinder in the first quadrant in the xy-plane given by $r(\theta)=2sin\theta$ for $0\leq \theta \leq \pi$ and dS is the areaelement across the curve with height z and width ds.
Then $ds=\sqrt{r(\theta)^2+(\frac{dr(\theta)}{d\theta})^2}=2d\theta$
So
The surface area you have found is absolutely correct. The book solution has an error. It defines curve $C$ as the curve of the cylinder in the first quadrant in the xy-plane. Now as the cylinder is in both first and second quadrants and the region is symmetric wrt y axis in xy-plane, it multiplies the answer by $2$. So it says -
$ \displaystyle \begin{equation} \iint_{S_1}\,dxdy = 2\int_{C}1\,dS \end{equation}=2\int_{C} |z| \,ds$
But if $C$ is only in the first quadrant in xy-plane, $0 \leq \theta \leq \frac{\pi}{2}$ and NOT $0 \leq \theta \leq \pi$.
Here are some more details that may explain to you the book solution. Your parametrization translates (no rotation) the coordinate axes such that to align the origin to the point $(0, 1)$ in the original coordinate system. The book uses the existing coordinates in which the cylinder is only in the first and second quadrants as seen in xy plane, and the parametrization uses polar coordinates $x = r \cos\theta, y = r\sin\theta, 0 \leq \theta \leq \pi$.
The equation of cylinder can be rewritten as $x^2 + y^2 = 2y \implies r = 2 \sin \theta, 0 \leq \theta \leq \pi$ in polar coordinates.
The equation of the cone is $z = - \sqrt{x^2+y^2} = - r$.
Now using the arc length formula in polar coordinates,
$ \displaystyle ds = \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} ~ d\theta$
As $r = 2 \sin\theta, ds = 2 ~ d\theta$
So the integral to find surface area $ \displaystyle \int_C |z| ~ ds$ can be rewritten as
$ \displaystyle \int_0^{\pi} 4 \sin\theta ~ d\theta = 8$
Or, $ ~ \displaystyle \int_0^{\pi/2} 8 \sin\theta ~ d\theta = 8$