Thread: 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?

The normal is just the gradient at some point on the surface of the equation so that's easy right:



But we want the point to be at the intersection of the equations and . 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 and since we want , then . Then the yellow contour is:

.

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



So I let , calculate that normal, then translate it to . 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}}]