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anonymous
 5 years ago
Consider the given curves to do the following.
y=x^3
y=0
x=1
Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = 1.
anonymous
 5 years ago
Consider the given curves to do the following. y=x^3 y=0 x=1 Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = 1.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would find the integral of x^3 from 0 to 1, then multiply that by by pi. It seems to give the same answer without going through a complicated integral set. It's been a while since I've touched one if these things, but in that rotation the radius is going to change values between the curve bounds. In the technique you are looking for it's not that it's impossible to solve but more complicated and less practical in the grand scheme of things. Since the radius of the rotation is one, why not just fnd that area and multiply it by a circle. Unless y9ou want to do \[\pi*\int\limits_{0}^{1}[1\sqrt[3]{y}]^2 dy\]. In that way it is like taking a cylinder and subtracting whatever circular shape comes from rotating the cubed root of y in a circle to get the volume. The whole idea in this whole series of techniques is to pick out how to get changes in radii in circular rotations. It pays whenever possible to take the easiest way, and having the knowledge to know you can.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y=x^3 ; inverse is: x = y^(1/3) y=0 x=1 ; > x=1 spin around: y =1. since we spin around a y; we inetgrate with respect to y; use equations in terms of x=______; and move the 2pi {S} y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02pi {S} (yR  yr) dy ; [1,0] 2pi {S} y(R  r) dy ; [1,0]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02pi  {S} y dy  {S} y(y 1)^(1/3)) dy  ; [1,0] integrate by parts that 2nd one: \[\frac{3y\sqrt[3]{(y1)^4}}{4}+\frac{9\sqrt[3]{(y1)^7}}{28}\] pi y^2 becomes: pi
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