Area of a Surface Defined Parametrically

If a surface is defined para metrically,  
\[\mathbf{r} (u,v)= x(u,v) \mathbf{i} + y(u,v) \mathbf{j} + z(u,v) \mathbf{k}(\]
  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  
\[\mathbf{r} = cos u \mathbf{i} + sin u \mathbf{j} +v \mathbf{k}\]
&nbsp with  
\[0 \leq u  
\[\frac|{\partial \mathbf{r}}{\partial u} = - sin \mathbf{i} + cos u \mathbf{j}, \frac|{\partial \mathbf{r}}{\partial v=1}\]

Then  
\[| \frac{\partial \mathbf{r}}{\partial u} du \times \frac{\partial \mathbf{r}}{\partial v} dv| = |(- sin \mathbf{i} + cos u \mathbf{j}) \times \mathbf{k} | = | cos u \mathbf{i} + sinu \mathbf{j} | = \sqrt{cos^2 u + sin^2 u} =1 \]

Then  
\[A= \int^{2 \pi}_0 \int^1_0 1 dv du = 2 \pi\]
 

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