Question

Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis.

y=tanx, 0<=x<=pi/4, about the x axis

Answer 1

Using "thin" washers parallel to the y-axis, thickness dx, then the region bounded by these lines {y = tanx , y = 0 , x= pi/4} in the interval for {x| 0 ≤ x ≤ (π/4)} , rotated around the axis y= -1 generates this volume:

integral from 0 to π/4 of π { [tanx + 1]² - 1} I ended up with π(0.6931472 ≈ 2.1776

When "triggering" with the quantity in the braces { } I came up with { [sec² x] + (2 tan x) } to integrate.