The region bounded by the x-axis and the curve y=2x-x^2 is rotated about the y-axis. the volume of the solid obtained=
You take the second integral right? then what do you plug in to get a number?
I have the second integral which should be (x^3/3)-(x^4/12)
2007-12-08 11:27:37 · 3 answers · asked by bitty_bri 2 in Science & Mathematics ➔ Mathematics
8pi/3 Thanks!
2007-12-08 11:59:37 · update #1
Volumes of revolution are calculated using either the method of disks/washers or the method of (cylindrical) shells. The latter method is the easier one to use for this problem.
The integral you should end up with, using the shells method, is
integral from 0 to 2 of 2πx(2x - x²) dx
2007-12-08 11:55:11 · answer #1 · answered by Ron W 7 · 0⤊ 0⤋
"...which of right here..." have been you meant to checklist some selection right here? Oh nicely...the radius of the forged is a million+sin^2(x), the barriers of integration are x=0 to x=pie/2. Now you're arising an limitless form of cylinders, infinitely narrow, whose quantity is section * top. you have the radius and additionally you have the top delta x. So combine the formulation for the quantity of a cylinder from 0 to pie/2, i.e., (pie)((a million+sin^2(x))^2)(delta x).
2016-11-14 22:49:21 · answer #2 · answered by Anonymous · 0⤊ 0⤋
NO...think 2 pi radius height thickness....x in[0,2]...the shell problem...not a 2 variable problem unless your 1st integral generate the surface area of the cylinder...and remember height is a measurement....close to 8.4
2007-12-08 11:49:50 · answer #3 · answered by ted s 7 · 0⤊ 0⤋