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Flux Out of a Parametrized Syrface - Example

Consider the surface defned by parameters  
\[u,v, 0 \leq u,v \leq 1\]
  so that  
\[\mathbf{r} = (u^2 +v^2)\mathbf{i} + uv \mathbf{j} + u \mathbf{k}\]
  , and the vector field  
\[\mathbf{F} = z \mathbf{i} + zy \mathbf{k}\]

The flux out of the surface is  
\[\int_S \mathbf{F} \cdot d \mathbf{S} \]

Because the surface is parametrized we can take
\[\begin{equation} \begin{aligned} \mathbf{F} \cdot d \mathbf{S} &= \mathbf{F} \cdot (\frac{\partial d \mathbf{r}}{\partial u} \times \frac{\partial d \mathbf{r}}{\partial v}) \\ &=(\mathbf{F} = z \mathbf{i} + zy \mathbf{k}) \cdot ( (2u \mathbf{i} + v \mathbf{j} + \mathbf{k}) \times (2v \mathbf{i} + u \mathbf{j})) \\ &= \left| \begin{array}{ccc} u & 0 & u^2 v \\ 2u & v & 1 \\ v & u & 0 \end{array} \right| \\ &= -u^2 +2u^2 v -2u^2 v^3 \end{aligned} \end{equation} \]

Then
\[\begin{equation} \begin{aligned} \int_S \mathbf{F} \cdot d \mathbf{S} &= \int^1_0 \int^1_0 -u^2 +2u^4 v -2u^2 v^3 du dv \\ &= \int^1_0 [- \frac{u^3}{3} + \frac{2u^5 v}{5} - \frac{2u^3 v^3}{3}]^1_0 dv \\ &= \int^1_0 - \frac{1}{3} + \frac{2 v}{5} - \frac{2 v^3}{3} dv \\ &=[- \frac{v}{3} + \frac{v^2}{5} -\frac{v^4}{6}]^1_0 \\ &= - \frac{1}{3} + \frac{1}{5} - \frac{1}{6} = - \frac{3}{10} \end{aligned} \end{equation}\]

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