Cylindrical Shell method?

Question

Use Cylindrical Shell method to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-axis.

y=x^3, x=1, y=0

please interns of dy

Answer 1

i hate these but no one else is doing it so i will answer.

2*pi*integral (x*x^3 dx) from [0,1]

2*pi [(x^5)/5] from [0,1]

and we get 2*pi

Answer 2

The cylindrical shell method involves approximating the volume under the surface of revolution by a series of concentric cylindrical shells, where the height of a particular shell between the values x1 and x1 + d is f(x1) and thus the volume contained by the shell is Pi*((x1 + d)^2)*f(x1) - Pi*((x1)^2)*f(x1 + d) = Pi*f(x1)*[(x1 + d)^2 - (x1)^2] = Pi*f(x1)*(x1 + d - x1)(x1 + d + x1) = Pi*f(x1)*(2x1 + d)*d. We uniformly slice the region over (x1, x2) into n cylindrical shells such that each shell has d = (x2 - x1)/n and sum these cylindrical slices over the interval (x1, x2), then take the limit as n increases without bound (or likewise, as d approaches 0) to get the integral form 2*Pi*Int[x1, x2][x*f(x)] dx.
In your case, we get the form 2*Pi*Int[0, 1][x^4] dx.Note the similarity to the formula C = 2*Pi*r, as if we were summing infinitesimally thin cylindrical shells of height f(x) and radius x.
There is no dy in the cylindrical shell method for this curve; you would use dy if you were using the disc method. In that case, you would rewrite the function as f(y) = y^(1/3) and use the integral form Pi*Int[0, 1][(y^(1/3))^2] dy, as if we were summing infinitesimally thin cylinders of radius f(y) and height dy.

Answer 3

before each and every thing; why do you want to sparkling up this in dy, at the same time as it truly is way a lot less annoying to sparkling up for dx? it truly is between the benefits of cylindrical shells; they're many times more effective for rotating around the y-axis bear in mind that the quantity is solved utilizing good the following formula. y=x^3, x = a million, y = 0 First, we state our go section. A(x) = 2pi(r)(h) = 2pi(x)(y) rationalization: we are attempting to sparkling up for the go section, when you consider that V = fundamental (A(x) dx) The go section is measured by the circumference of between the shells circumstances the top. x represents the radius, and y represents the top. even with the undeniable fact that, y = x^3, so A(x) = 2pi(x)(x^3) So our fundamental will change into fundamental (2pi(x)(x^3)dx OR, pulling out the consistent, 2pi * fundamental (x(x^3))dx 2pi * fundamental (x^4) dx What are our bounds of integration? For one element, we are fixing this problem contained in the x-route (our shells are stacked from left to good). the first certain will easily be the y-axis, 0. the 2d certain is x = a million. hence, we stumble on the fundamental from 0 to at least a million. 2pi * fundamental (0 to at least a million, x^4) dx which supplies us 2pi * [ (x^5)/5, evaluated from 0 to at least a million ] 2pi * [a million/5] 2pi/5