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 Jan25 comment Sense of rotation. How would the rotation matrix look like for this “arbitrary” axis? Yes. See the arrow in the picture showing the direction of the rotation axis. Jan25 answered Sense of rotation. How would the rotation matrix look like for this “arbitrary” axis? Jan20 answered What are the degrees of freedom of $F=ma$ and $F =mdv/dt$? Jan13 comment Factors of the RSA modulus Jan12 answered Using the chain rule backwards Jan4 comment Arc length of a 3D Curve If you want higher precision, you create a cubic spline from the points and integrate the spline. Jan4 answered RK4 method applied to $\frac{dy}{dt}=-\frac{y-t}{2}$ with $y(0)=1$ Jan4 comment RK4 method applied to $\frac{dy}{dt}=-\frac{y-t}{2}$ with $y(0)=1$ Why didn't you try it with RK4? Juts Google it if you are lost. Jan1 comment solving systems of equations for m and b when you know they are both positive? If you expected a positive result and didn't get it, then there is something wrong in your model (the equations or the solution). Maybe a coefficient or a sign somewhere. I get $m=1125$ and $b=88700$. I suggest showing how you get your values. Dec15 awarded Caucus Nov24 comment Midpoint of the shortest distance between 2 rays in 3D The vector from $a$ to $p$ projected onto the line is $b\left(\frac{b\cdot (p-a)}{b\cdot b}\right)$ and the above is another way of doing projections using the vector triple cross product. Nov24 revised Midpoint of the shortest distance between 2 rays in 3D added 491 characters in body Nov24 answered Midpoint of the shortest distance between 2 rays in 3D Nov19 comment Ratio of Areas of Similar Triangles The image link is broken for me. Nov10 comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$ The difficulty is the for a thin shell $\int dx = 0$, but $\int \rho dx >0$. Real mass density is infinite since the material volume of a hollow sphere is zero. Nov10 comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$ If real mass density $\rho$ is used then $w = \int \rho {\rm d}x$ Now the limits of this integration depend on $y$. To do this right your slicing has to be done with a polar angle ${\rm d}\theta$ instead of ${\rm d}y$. Then $dm = \rho \pi x^2 R d \theta d R$ instead of $dm=\rho \pi x^2 dx dy$ Nov10 comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$ Dimensionally you integral is incorrect. The result should be of the mass times distance squared units. Nov10 answered Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$ Nov10 comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$ I think your $dm$ is suspeect. If you do $\int dm = \int \frac{M \sqrt{R^2-y^2}}{2 R^2}\,{\rm d}y = \frac{\pi}{8} M R \ne M$. Shouldn't by definition $M=\int dm$? Nov10 comment Evaluate ${M\over2R^2}\int(R^2 - y^2)^{3\over2}dy$ Moment of inertia calculation should not have a fractional exponent. Can you show more of how you arrived at this.