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 interms of dy

Answer 1

X=Y^1/3 , X FUNCTION ON Y
FOR X=1 THEN Y=1
the volume is the antiderivate of function (y^1/3)^2 for y=0 to1.
the volume is (pi*y^5/3)/(5/3) for y=0 to1
the volume is 3pi/5.

Answer 2

The cylindrical shell approach contains approximating the quantity lower than the outdoors of revolution by a sequence of concentric cylindrical shells, the position the top of a particular shell between the values x1 and x1 + d is f(x1) and therefore the quantity 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 area over (x1, x2) into n cylindrical shells such that each and each and every shell has d = (x2 - x1)/n and sum those cylindrical slices over the period (x1, x2), then take the reduce as n will boost with out certain (or likewise, as d procedures 0) to get the fundamental type 2*Pi*Int[x1, x2][x*f(x)] dx. on your case, we get the type 2*Pi*Int[0, a million][x^4] dx.note the similarity to the formula C = 2*Pi*r, as if we were summing infinitesimally skinny cylindrical shells of top f(x) and radius x. there isn't any dy contained in the cylindrical shell approach for this curve; you would use dy in case you've been utilizing the disc approach. if so, you would rewrite the function as f(y) = y^(a million/3) and use the fundamental type Pi*Int[0, a million][(y^(a million/3))^2] dy, as if we were summing infinitesimally skinny cylinders of radius f(y) and top dy.

Answer 3

First off; why do you want to solve this in dy, when it is much easier to solve for dx? That's one of the advantages of cylindrical shells; they're usually better for rotating around the y-axis

Recall that the volume is solved using the following formula.

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

First, we state our cross section.

A(x) = 2pi(r)(h) = 2pi(x)(y)

Explanation: We're trying to solve for the cross section, since

V = Integral (A(x) dx)

The cross section is measured by the circumference of one of the shells times the height. x represents the radius, and y represents the height.

However, y = x^3, so

A(x) = 2pi(x)(x^3)

So our integral becomes

Integral (2pi(x)(x^3)dx

OR, pulling out the constant,

2pi * Integral (x(x^3))dx
2pi * Integral (x^4) dx

What are our bounds of integration? For one thing, we're solving this problem in the x-direction (our shells are stacked from left to right). The first bound will definitely be the y-axis, 0. The second bound is x = 1. Therefore, we find the integral from 0 to 1.

2pi * Integral (0 to 1, x^4) dx

Which gives us

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