# What have I done?

1. Nov 9, 2011

### ShayanJ

I know my question is strange(and maybe stupid) but I'm really curious about it.I once tried to calculate the gradient of the magnitude of the gradient of a scalar field which for its x coordinate,I found:

$\left[ \nabla \left( \left| \nabla \varphi \right| \right) \right]_{x} = \frac {\frac { \partial ^ {2} \varphi } {\partial x ^ {2} } } {\sqrt { 1+ (\frac{\frac {\partial \varphi} {\partial y} } {\frac {\partial \varphi } {\partial x} } )^{2} } }$

then I tried to write it in a simpler form and it just came into my mind that I can integrate it and write it as the derivative of a function so I made substitutions below and treated $\frac {\partial \varphi} {\partial y}$ as constant:

$\frac{\frac {\partial \varphi } {\partial y} }{\frac {\partial \varphi } {\partial x} }=\tan{\theta}$
$\frac{\partial ^{2} \varphi }{\partial x ^{2} }=- \frac {\partial \varphi } {\partial y} \csc ^{2}{\theta} d \theta$

so I arrived at:

$\int \left[ \nabla \left( \left| \nabla \varphi \right| \right) \right]_{x} = \frac {\partial \varphi} {\partial y} \csc {\theta}$

I reverted and got the following:

$\int \left[ \nabla \left( \left| \nabla \varphi \right| \right) \right]_{x} = \left| \nabla \varphi \right|$

Now comes my question:
When I check the things I've done,I just can tell bullsh*t.But the result tells that was really an inegration but I really don't understand how can that be an integration.

Sorry for such a mess and thanks in advance

Last edited: Nov 9, 2011
2. Nov 9, 2011

### dalcde

I'm not really sure what you did (I'm not really good at that stuff), but I don't think you have to treat $\frac {\partial \varphi} {\partial y}$ as constant since $\theta$ can be a variable (I'm not really sure if I'm right, but I think it works that way).