Michiel
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 Jul 28 comment Zero-order interpolation not going fully around Terrific summary of the problem and our discussion in the comments. Thanks again Jul 26 comment Extrude 2D cross-section to 3D shape with shrinkfactor This is indeed the route that I explored at first, but it is not entirely what I was looking for, because this remains a body of slices instead of a complete surface. Jul 26 comment Extrude 2D cross-section to 3D shape with shrinkfactor Thanks a bunch for the thorough answer, it really makes for beautiful shapes now!! Jul 26 comment Extrude 2D cross-section to 3D shape with shrinkfactor Thanks, but never mind. I already figured out the issue (see my edit in the question). Of course, the error was due to an adaptation that I made to your code, so I've only got myself to blame. Jul 26 comment Extrude 2D cross-section to 3D shape with shrinkfactor This works brilliantly, except for one thing that I cannot figure out: the cross-section for my own data is scaled to half the size in the extruded shape?! If I put a factor 2 inside the append (like Append[2*thicknessFunc[u,2]*fdata[t], u]) then it is fixed, but I really don't understand. I will add this to my question in an edit Jul 4 comment Forcing an integral to be solved in separate terms Great, thanks! Quick comment: did you know that you can copy the output of mathemetica directly as latex by right-clicking on the equation and selecting "Copy as -> Latex"? Jan 20 comment Count the number of sqrt and 4th powers in a function Nice, thanks! A small follow-up: how does {0, Infinity} differ from just Infinity? Because if I use for example Count[Sec[x], Sec[__],..] mathematica returns 0 if I only use Infinity, but 1 if I use the {0,Infinity}' Jan 15 comment How to find the most compact form of an equation So if I understand correctly I can in theory give any ComplexityFunction I like?! Do you know what happens when I give a very simple ComplexityFunction e.g. the number of times a certain variable appears in the equation, and there are multiple equal solutions. Does it then use its default ComplexityFunction to distinguish between those or is one of the options 'randomly' given as best solution? Jan 15 comment How to find the most compact form of an equation @SjoerdC.deVries I did indeed specify assumptions on variables being both real and positive. Without that the original equation would have been several lines longer. The ComplexityFunction seems to be what I am looking for. Mar 22 comment Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ I had tried it, but it gave a very different result. However, looking at the documentation I found that I should use ArcTan[b^2 Cos[phi]],a^2 Sin[phi]] to get the correct result. Thanks! If you could write it as an answer I will definitely accept it. Mar 22 comment Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ @MichaelE2 no I am sure the phase-shift is the correct thing to do. The reason is that ArcTan[(a/b)^2 Tan[[Phi]]] represents the angle the interface normal of an ellipse makes with the x-axis as determined from the angle the local x,y coordinates make with the x-axis (this angle is $\phi$). So I know that the whole thing should run from 0 to 360 deg and that the minima and maxima should coincide with those found when a=b. ---- So maybe I didn't explain enough of the background of the problem in my original question, but this is what I was looking for Mar 21 comment Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ Thanks for the explanation of how to get the answer to the integral, but I found out that it was in fact not the answer to the integral that was incorrect, but rather what I fed it: the Cos[ArcTan[ Tan[[Phi]]]] which has a modulus at $\pi/2$ Feb 27 comment Stop Mathematica from giving imaginary solutions @DanielLichtblau thanks for the explanation, I was not aware of that phenomenon. The Cubics->False option doesn't solve it however Feb 26 comment Stop Mathematica from giving imaginary solutions I have done what you mentioned, except for the //First//First part, because that will give me the wrong 1 of the 3 answers. Feb 26 comment Stop Mathematica from giving imaginary solutions Never mind this comment..... Too quick to respond while you where still explaining Feb 26 comment Stop Mathematica from giving imaginary solutions That is actually a good point. I didn't know that Solve had that option. I tried using \$Assumptions, but that didn't help. Thanks! Feb 26 comment Stop Mathematica from giving imaginary solutions Not exactly. I found this post but I would like Solve to work with the assumptions already 'in mind'. Not change' the answer afterwards