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 Mar8 comment How to sketch the curve of parametric equations sorry i had to finish editing my response Mar8 comment How to sketch the curve of parametric equations i see now. so sin θ >= 0 and then sin θ <= 1. so x >= 0 and <=1? Mar8 comment How to sketch the curve of parametric equations oh ok. so i could create a table of various points for t and then find the corresponding x and y values to determine the curve? Mar8 comment How to sketch the curve of parametric equations oh ok. so essentially, you can just plot the initial point, the ending point to get the shape of the curve, and a point in between to determine the direction of the curve? Mar8 comment Find the area of the surface obtained by rotating the curve about the x axis ahh now i see! thanks Mar7 comment Find the area of the surface obtained by rotating the curve about the x axis After doing that I get: the integral from 0 to 1 of [(-2/pi) sqr(1 + u^2pi) du]. Would I have to do a trig substitution? Mar3 comment Determining a closed-form solution for the following sum thanks! that cleared it up Mar3 comment Determining a closed-form solution for the following sum but when evaluating the first summation, it does not need to be multiplied by (n+1) because you are plugging in for i directly? Mar3 comment Determining a closed-form solution for the following sum in the second summation, are you adding n+1 because the index starts at 0 and not 1? Mar3 comment Determining a closed-form solution for the following sum the second one. Feb27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ great. thanks for your help! Feb27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ Thanks. I finally understand, but shouldn't the evaluated integral be: -1/(10x^3) + x^5/6? Feb27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ how did you get the second qauntity exactly? Feb27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ is there supposed to be a sqr root? Feb27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ @Tavares, yes it is Feb27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ ok now i see. so then the integral would be sqr[(25x^8)/36 + 1/2 + 9/(100x^8)] Feb27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ @Tavares where does the 1/2 come from? Feb27 comment Find the length of the curve: $y=\frac{x^{5}}{6}+\frac{1}{10x^{3}}\qquad 1\leq x\leq 2$ It is 1/(10x^3) Feb25 comment Number Theory - Proof of divisibility by $3$ ahh i see now. makes perfect sense. thanks Feb25 comment Number Theory - Proof of divisibility by $3$ @yunone, how could you prove that 3|x by using x=y(mod 3) if 3|(x-y) only