# michielm

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bio website tudelft.nl location Delft, Netherlands age 25 member for 3 months seen May 5 at 11:23 profile views 4

I am a PhD student in the field of fluid dynamics and soft matter.

# 24 Actions

 Apr11 awarded Teacher Mar26 awarded Supporter Mar23 revised Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$edited body Mar22 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. Mar22 revised Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$added 258 characters in body Mar22 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 Mar21 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 $\pi/2$ Mar21 revised Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$added 4 characters in body Mar21 answered Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ Mar21 revised Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$added 183 characters in body Mar21 revised Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$some additional information Mar21 asked Evaluating the integral of $\cos (\theta )\cos\left[\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)\right]$ Mar2 awarded Autobiographer Feb27 awarded Scholar Feb27 accepted Stop Mathematica from giving imaginary solutions Feb27 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 Feb26 awarded Editor Feb26 comment Stop Mathematica from giving imaginary solutionsI have done what you mentioned, except for the //First//First part, because that will give me the wrong 1 of the 3 answers. Feb26 revised Stop Mathematica from giving imaginary solutionsA step further in the question Feb26 awarded Student