how to determine the outward pointing normal (gauss divergence theorem)

I have a cone defined by $x^2+y^2=(1-z)^2$ i was trying to work out the normal vector on surface $s_1$ indicated on the plot On $s_1$: r=$\left<x,y,0\right>$ since $z=0$ on $x-y$ plane

$\dfrac{\partial}{\partial x}=1;$ $\dfrac{\partial}{\partial y}=1;$ $\dfrac{\partial}{\partial z}=0$

$\therefore$ n=$\dfrac{\partial}{\partial x}$×$\dfrac{\partial}{\partial y}$ =$\left<0,0,1\right>$, however my textbook "thinks" on $s_1$ this normal vector points inwards and the outward pointing normal is given by n=$\dfrac{\partial}{\partial x}$×$\dfrac{\partial}{\partial y}$ =$\left<0,0,-1\right>$. Can someone please explain why this is, how do i know where the normal vector is pointing -Thanks.

EDIT:

i also dont't understand why this is wrong for $s_2$:

$r$=$\left<x,y,1-\sqrt{x^2+y^2}\right>$ note that $x^2+y^2=(1-z^2)$ for this question:

$r_x$=$\left<1,0,\dfrac{-x}{\sqrt{x^2+y^2}}\right>$ and $r_y$=$\left<0,1,\dfrac{-y}{\sqrt{x^2+y^2}}\right>$

$r_x$×$r_y$ =$\left<\dfrac{x}{\sqrt{{x^2+y^2}}}, \dfrac{y}{\sqrt{{x^2+y^2}}},1 \right>$

$\implies n$=$\left<\dfrac{-x}{\sqrt{{x^2+y^2}}}, \dfrac{-y}{\sqrt{{x^2+y^2}}},-1 \right>$

shouldn't $-\left<\hat{i},\hat{j},\hat{k}\right>$ point away from the surface at $s_1$?

• hmmm, so are you saying that an outward pointing normal is one such that it point away from the orientation of $z$ i.e. if $z$ is positive then normal should be negative? – Ozwurld Dec 30 '14 at 22:03
• i understand what you are saying, however i dont understand why this is wrong for $s_1$: $r$=$\left<x,y,1-\sqrt{x^2+y^2}\right>$ note that $x^2+y^2=(1-z^2)$ for this question: $r_x$=$\left<1,0,\dfrac{-x}{\sqrt{x^2+y^2}}\right>$ and $r_y$=$\left<0,1,\dfrac{-y}{\sqrt{x^2+y^2}}\right>$ $r_x$×$r_y$ =$\left<\dfrac{x}{\sqrt{{x^2+y^2}}}, \dfrac{y}{\sqrt{{x^2+y^2}}},1 \right>$ $\implies n$=$\left<\dfrac{-x}{\sqrt{{x^2+y^2}}}, \dfrac{-y}{\sqrt{{x^2+y^2}}},-1 \right>$ shouldn't $-\left<\hat{i},\hat{j},\hat{k}\right>$ point away from the surface at $s_1$? – Ozwurld Dec 31 '14 at 5:27
• The above is for $s_2$ not $s_1$ – Ozwurld Dec 31 '14 at 6:29
• @Ozwurld The normal for $S_2$ should have positive z coordinate if you want it to point outward (from the object). – Ofir Dec 31 '14 at 11:29