\[\mathbf{r} (u,v)= x(u,v) \mathbf{i} + y(u,v) \mathbf{j} + z(u,v) \mathbf{k}(\]
then we can define an element of surface area \[dA = | \frac{\partial \mathbf{r}}{\partial u} du \times \frac{\partial \mathbf{r}}{\partial v} dv| \]
and the surface area of the surface is\[S =\int_u \int_v | \frac{\partial \mathbf{r}}{\partial u} du \times \frac{\partial \mathbf{r}}{\partial v} dv| dv du\]
The surface of a cylinder can be defined para metrically as
\[\mathbf{r} = cos u \mathbf{i} + sin u \mathbf{j} +v \mathbf{k}\]
  with \[0 \leq u Then {jatex options:inline}| \frac{\partial \mathbf{r}}{\partial u} du \times \frac{\partial \mathbf{r}}{\partial v} dv| = |(- sin \mathbf{i} + cos u \mathbf{j}) \times \mathbf{k} | = | cos u \mathbf{i} + sinu \mathbf{j} | = \sqrt{cos^2 u + sin^2 u} =1 \]
Then
\[A= \int^{2 \pi}_0 \int^1_0 1 dv du = 2 \pi\]