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4

This is documented behavior. The parametric forms of ArcLength, Area, and Volume take the parametrization as fundamental, and compute the area including multiple coverings. The region forms of the functions take the image as fundamental, and compute the area of the embedding into R^n. If you want the latter, you should use ParametricRegion. From the ...


3

I think you should step back and think carefully about the expression you're using. Why are you using Sin[p] to represent the height? Also, if the height is 2, then it is not 2π. expr = {Sin[t], Cos[t], p}; ParametricPlot3D[ expr, {t, 0, 2 \[Pi]}, {p, 0, 2}, AxesLabel -> (Style[#, 18, "Label", Blue] & /@ {"x", "y", "p"}) ] Area[expr, {t, 0, 2 ...


6

Let mi visualize this mistake. Instead of constant radius we will use radius that depends of p: expr = {1 + .1 p, 1 + .1 p, 1} { Sin[t], Cos[t], Sin[p]}; ParametricPlot3D[expr, {t, 0, 3 Pi/2}, {p, 0, 2 \[Pi]}] As you can see, for p limits: {0, 2Pi} you are plotting your surface twice. Try with Cos[p] and p in {0,Pi} or whatever monotonic function for ...


6

For this particular example versio 10 functionality is helpful: list = {{0, 13}, {8, 10}, {13, 6}, {10, 0}}; pg = {{0, 0}}~Join~list; rm[x_, y_] := RegionMember[Polygon[pg], {x, y}] You can see criteria: Reduce[rm[x, y]] yielding: (0 <= y <= 6 && 0 <= x <= (20 + y)/2) || (6 < y <= 10 && 0 <= x <= 1/4 (82 - 5 ...


4

Try this: list = {{0, 13}, {8, 10}, {13, 6}, {10, 0}}; A = {5, 3}; f = Interpolation[list, InterpolationOrder -> 1]; If[f[A[[1]]] < A[[2]], "Over the curve", "Under The curve"] (*"Under The curve"*) Based on your comment, you are looking to check if the point is enclosed within the curve and the line y = 0. You can try this: list2 = Join[{{0, ...


0

I found the answer by Szabolcs really useful in finding the answer to the same question, so I thought I'd just leave the exact way to do it here in case someone stumbles across it like I did. The speed boost is very good - I'd been relying on this function to get the output I needed and for a list of about 3000 points it was taking about 20 seconds on my ...


0

No answer here but only further forward suggestions with my thoughts on the topic. We can start with any contour C but more conveniently consider a loop with known closed form parametrization. Supposing we start with an "ellipse" contour C written on a unit sphere ( defined by achille hui in SE Math in reply to my question or any Monkey saddle variant) with ...



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