I am having a bit of trouble understanding example 1 in Paul's Calculus Notes page on surface integral:
http://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx
I understand how to do surface integrals by parametrizing the surface; however, in the problem, the author directly calculates the normal vector by finding the gradient and simply ends up taking the dot product between the vector field and the normal vector.
I understand how this may work if the normal vector is normalized to a unit vector, and he does divide by the magnitude; however, I don't understand why he is multiplying by the magnitude of the normal vector again and saying that the magnitude will "cancel out."
Thank you.
The surface area element of the graph of a function $ z=f\,(x,y)$ is
$$dA= \left\|{\partial \mathbf{r} \over \partial x}\times {\partial \mathbf{r} \over \partial y}\right\| \mathrm dx\, \mathrm dy $$
where $\mathbf{r}=(x, y, z)=(x, y, f(x,y))$. So that ${\partial \mathbf{r} \over \partial x}=(1, 0, f_x(x,y))$, and ${\partial \mathbf{r} \over \partial y}=(0, 1, f_y(x,y))$. Then,
$$dA = \left\|\left(1, 0, {\partial f \over \partial x}\right)\times \left(0, 1, {\partial f \over \partial y}\right)\right\| \mathrm dx\, \mathrm dy \\ =\left\|\left(-{\partial f \over \partial x}, -{\partial f \over \partial y}, 1\right)\right\| \mathrm dx\, \mathrm dy \\ =\sqrt{\left({\partial f \over \partial x}\right)^2+\left({\partial f \over \partial y}\right)^2+1}\, \, \mathrm dx\, \mathrm dy$$
(see Wikipedia)