Partial Derivative of the conical surface $z=\sqrt{x^2+y^2}$

Question

The example in my textbook is attempting to calculate the surface integral of $\int\int_Sz\,dS$ where $z=\sqrt{x^2+y^2}$ between z=0 and z=1. They state that since $z^2=x^2+y^2$ that $\frac{\partial z}{\partial x}=\frac{x}{z}$ and that $\frac{\partial z}{\partial y}=\frac{y}{z}$. I have no clue where these are coming from.

Answer 1

Derive the expression $z^2=x^2+y^2$ with respect to $x$ and use the chain rule to get $$2z \frac{\partial z}{\partial x} = 2x+0.$$ Hence $$\frac{\partial z}{\partial x}=\frac{x}{z}.$$

Answer 2

The usual advice if you don't understand what happened in some passage in your textbook: just try to do it your own way instead.

Like this, perhaps: $$ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \sqrt{x^2+y^2} = \frac{1}{2 \sqrt{x^2+y^2}} \cdot 2x = \frac{x}{\sqrt{x^2+y^2}} . $$ Nothing strange so far!

But wait, now we notice that the expression downstairs is just $z=\sqrt{x^2+y^2}$, so we could write the whole thing as $$ \frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{z} , $$ if we wanted to. That's it, mystery solved.

(That said, what the the authors probably intended is indeed the calculation given in the answer by C. Dubussy.)