Recent content by Tarpie

ya...because each point on that arc length is mapped to a certain x value regardless of whether in terms of y or x...so you can just use that x and the x interval instead of flipping it and using the y interval in terms just because it's about y...like rotation of height only sideways...anyways...

I would suspect it's just surface area because of that invariant arc length term

Does this apply to volume as well or just surface area, because for volumes I've always rearranged for x as a function of y if it asks to rotate a curve about the y axis.

I see now. That x in your equation I'm used to thinking of as f(y), just like when you rotate the curve about the x-axis it's a y; i also see that as an f(x). So for each radius which is each x value on the curve i thought I had to rewrite it as a function of y integrate over the y...

Sorry y-axis. Could've sworn i wrote it

Homework Statement [/B] Find the surface area obtained by rotating the curve y = x^2/4 - ln(x)/2 1 \leq x \leq 2 Homework Equations 2π \int f(x)\ \sqrt{1+(f'x)^2} dx The Attempt at a Solution I can't seem to isolate for x in terms of y. I raised both sides to e and separated the exponents...

Hey I don't mean to hijack but let me just add to this inquiry. If a function is equal to its mclaurin series and the mclaurin series has a radius of convergence of infinity, then the function is equal to to that series at all X right? ex equal to it's mclaurin series ∑Xn/n! for every X. But...

Greetings, y=x2/4 - ln(x)/2 from 1=<x<=2 rotated about the y-axis. I did the equation rotating about the x-axis via 2pi* integral (f(x)*sqrt(1+f'(x)^2)) dx with dy/dx = x/2 - 1/2x but the question calls for rotation about y and i can't seem to rearrange the equation to isolate for...

Greetings everyone, I'm a mechanical engineering and mathematics student interested in all things physics and math, especially Newtonian mechanics, SR & GR/ analysis and geometry. Thank you and I hope to learn a lot here!