I think I can imagine the shape of surface area. This is what I did:
$$\begin{eqnarray*} \text{Surface area} & = & \int_{0}^{\pi} r y \sqrt {2} \, d\theta \\ & = & \int_{0}^{\pi} r\cdot r \sin \theta \sqrt {2}\, d\theta \end{eqnarray*}$$
But my answer is not correct. Where is my mistake? Thank you
The height of the surface is $r\cdot sin\phi$.
C is the half-circle $C_x(\phi) = rcos\phi$ $C_y(\phi) = rsin\phi$ $$S=\int_{C}z ds = \int_{0}^{\pi}r\cdot sin\phi\sqrt{\bigl(\frac{dC_x}{d\phi}\bigr)^2 + \bigl(\frac{dC_y}{d\phi}\bigr)^2}d\phi =$$ $$=\int_{0}^{\pi}r\cdot sin\phi\sqrt{r^2sin^2\phi + r^2cos^2\phi}d\phi= r^2\int_{0}^{\pi} sin\phi d\phi = r^2\bigl[-cos\phi\bigr]_{0}^{\pi} = 2r^2$$