Cylindrical shell method?

Question

use cylindrical shells to find the volume of the solid that results when the region enclosed by y^2=8x and x=2 is revolved about the line x=4.

I got (64(4sqrare root 2-1)pie)/5. Is it correct?

Answer 1

I get the same answer. I used 2*pi*integral(x*f(x)) where f(x) is sqrt(8x) and the limits are 2 to 4.

Answer 2

No trouble will be if you shift the whole stuff by 4 to the left so that y^2=8(x+4) and x+4=2 or y=s*sqrt(8x+32)), where s=-1 or +1, x=-2 and Y-axis is axis of rotation;
now the volume of elementary shell dv=-(2pi*x*dx) * h,
where h = sqrt(8(x+4))-(- sqrt(8(x+4)) or upper-branch minus lower-branch, -2pi*x*dx is the area of base of the elem.shell (minus to compensate negative x);
thus dv=-4pi*x*sqrt(8x+32)*dx, while –4 V=-4pi * integral{for x=-4 until –2} of x*sqrt(8x+32)*dx;
let u=sqrt(8x+32) then x=(1/8)*u^2 – 4,
dx=(1/4)*u*du {for u=0 until 4};
V=-pi*integral{for u=0 until 4}of ((1/8)*u^2 –4)*u^2*du =
= pi*[(1/8)*(1/5)*u^5 – (4/3)*u^3] = -pi*(4^5 / 40 – (4/3)*4^3) = -pi*64*(14/15) = 187.6578; while your answer = 187.2632; close but not the same ?! then check me!

Answer 3

There is symmetry about the x axis so you can do just the upper halve and double the answer:

I looked ahead and noticed an ambiguity: is x=2 the inner boundary or the exterior? The following assumes that the shape has a hole of radius=2.

a dV = 2*pi*r*h*dr

now, h and r are functions of x:

r= 4-x , h = ksqrt(x) where k=sqrt(8)

then integrate dV from x=0 to x=2.

Answer 4

The cylindrical shell technique is composed of approximating the volume lower than the exterior of revolution with techniques from a chain of concentric cylindrical shells, the position the properly of a particular shell between the values x1 and x1 + d is f(x1) and for that reason the volume contained with techniques from 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 decrease as n will strengthen without sure (or likewise, as d techniques 0) to get the fundamental style 2*Pi*Int[x1, x2][x*f(x)] dx. on your case, we get the style 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 properly f(x) and radius x. there's no dy interior the cylindrical shell technique for this curve; you should apply dy in case you've been utilising the disc technique. if so, you should rewrite the function as f(y) = y^(a million/3) and use the fundamental style Pi*Int[0, a million][(y^(a million/3))^2] dy, as if we were summing infinitesimally skinny cylinders of radius f(y) and correctly dy.