# Finding normal vector for line/surface integrals

• December 5th 2009, 10:32 PM
messianic
Finding normal vector for line/surface integrals
For instance, if curve C lying on the intersection of the cylinder x^2 + y^2 = 2y and the plane y = z, how would I find the normal vector n for this surface?

You often have to find the normal vector when evaluating a line or surface integral. I have trouble with this. Does anyone have any tips? Or a guide to go about finding it?
• December 6th 2009, 07:10 AM
shawsend
The normal is just the gradient at some point on the surface of the equation $F(x,y,z)=0$ so that's easy right:

$n=\nabla F]_{p_0}$

But we want the point $p_0$ to be at the intersection of the equations $F(x,y,z)=x^2+y^2-2y=0$ and $g(x,y)=z=y$. You can complete the square of the first one and see in the x-y plane that it's a circle with radius one at center (0,1). So parametrically, that's $x=\cos(t), y=1+\sin(t)$ and since we want $y=z$, then $z=1+\sin(t)$. Then the yellow contour is:

$\{\cos(t),1+\sin(t),1+sin(t)\}$.

So I can let any point $p_0$ be on that contour and then calculate the normal there:

$\nabla F]_{p_0}$

So I let $t=0$, calculate that normal, then translate it to $p_0$. That's the red one. Here's the Mathematica code I used in case you want to work with it:

Code:

p1 = ParametricPlot3D[{Cos[t], 1 + Sin[t], 1 + Sin[t]}, {t, 0,     2 \[Pi]}, PlotStyle -> {Thickness[0.008], Yellow}]; xpt = 1; ypt = 1; zpt = 1; myNormal = {2 x, 2 y - 2, 0} /. {x -> xpt, y -> ypt, z -> zpt}; myLine = Graphics3D[{Thickness[0.01], Red,     Line[{{xpt, ypt, zpt}, {xpt + myNormal[[1]], ypt + myNormal[[2]],       zpt + myNormal[[3]]}}]}]; cp1 = ContourPlot3D[{x^2 + y^2 == 2 y}, {x, -1, 2}, {y, -1,     3}, {z, -2, 3}]; cp2 = ContourPlot3D[{y == z}, {x, -1, 2}, {y, -1, 3}, {z, -2, 2},   ContourStyle -> Opacity[0.2]]; Show[{cp1, cp2, p1, myLine}, BoxRatios -> {1, 1, 1},  PlotRange -> {{-3, 3}, {-3, 3}, {-3, 3}}]