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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 plotplot

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 $s_1$?

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According to the picture (and the equation of the cone), the cone is above the XY plane (namely with positive z coordinate), so that the normal on its basis that points outward should be with a negative sign in the z coordinate. When using the divergence theorem you always need to choose the outward normal, so you first need to calculate a normal using the vector product and then multiply by -1 if needed.

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  • $\begingroup$ 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? $\endgroup$ – Ozwurld Dec 30 '14 at 22:03
  • $\begingroup$ Not exactly. An outward means that if you move a little bit in the direction of the vector, you will leave the object. It doesn't mean that the entire object is in the other direction (as it is in this case). For example consider a ball of radius 2 and then take out a ball of radius 1 and look at the normal outward (from the object) at point (1,0,0). $\endgroup$ – Ofir Dec 30 '14 at 22:10
  • $\begingroup$ 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 $s_1$? $\endgroup$ – Ozwurld Dec 31 '14 at 5:27
  • $\begingroup$ The above is for $s_2$ not $s_1$ $\endgroup$ – Ozwurld Dec 31 '14 at 6:29
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    $\begingroup$ @Ozwurld The normal for $S_2$ should have positive z coordinate if you want it to point outward (from the object). $\endgroup$ – Ofir Dec 31 '14 at 11:29

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