How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y =x^3, y= 8 , x= 0 revolved about the x-axis?

1 Answer
Feb 26, 2017

Volume = (768pi)/7 \ \ \ unit^3

Explanation:

If you imagine an almost infinitesimally thin vertical line of thickness deltax between the x-axis and the curve at some particular x-coordinate it would have an area:

delta A ~~"width" xx "height" = ydeltax = f(x)deltax
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If we then rotated this infinitesimally thin vertical line about Oy then we would get an infinitesimally thin cylinder (imagine a cross section through a tin can), then its volume delta V would be given by:

delta V~~ 2pi xx "radius" xx "thickness" = 2pixdeltaA=2pixf(x)deltax

If we add up all these infinitesimally thin cylinders then we would get the precise total volume V given by:

V=int_(x=a)^(x=b)2pi \ x \ f(x) \ dx

Similarly if we rotate about Ox instead of Oy then we get the volume as:

V=int_(y=a)^(y=b)2pi \ y \ g(y) \ dy

So for this problem we have:

enter image source here
We need the point of intersection for the bounds of integration;

x \ = 0 => y=0
x^3 = 8=>x=2

And we are integrating wrt y so we need x=g(y) (ie f^-1(x))

y=x^3 => x=y^(1/3)

Then the required volume is given by:

V=int_(y=a)^(y=b)2pi \ y \ g(y) \ dy
\ \ \= 2pi int_0^8 \ y*y^(1/3) \ dy
\ \ \= 2pi int_0^8 \ y^(4/3) \ dy
\ \ \= 2pi [y^(7/3)/(7/3)]_0^8
\ \ \= (6pi)/7 [y^(7/3)]_0^8
\ \ \= (6pi)/7 (8^(7/3)-0)
\ \ \= (6pi)/7 * 128
\ \ \= (768pi)/7 \ \ \ unit^3

Washer Method
We call also use the "washer" method which gives the volume of revolution about Ox (this is the non-shaded area under the curve) as:

V=pi \ int_(x=a)^(x=b) \ (f(x))^2 \ dx

Which in this problem gives:

V="(volume bounded by " y=8) - ("volume bounded by " y=x^3)

Volume bounded by y=x^3 is:

V=pi \ int_0^2 \ (x^3)^2 \ dx
\ \ \ =pi \ int_0^2 \ x^6 \ dx
\ \ \ =pi \ [x^7/7]_0^2
\ \ \ =pi \ (128/7 - 0)
\ \ \ =(128pi)/7

The required volume of revolution is then the volume of a cylinder (pir^2h) of height 2, radius 8 minus this result:

V=(pi)(64)(2) - (128pi)/7
\ \ \ =128pi - (128pi)/7
\ \ \ =(768pi)/7 , as before

Or if you prefer, we can perform the calculation in a single integral, as:

V="(volume bounded by " y=8) - ("volume bounded by " y=x^3)
\ \ \ =pi \ int_0^2 \ (8)^2 \ dx - pi \ int_0^2 \ (x^3)^2 \ dx
\ \ \ =pi \ int_0^2 \ (8)^2 - (x^3)^2 \ dx
\ \ \ =pi \ int_0^2 \ 64-x^6 dx
\ \ \ =pi \ [64x-x^7/7]_0^2
\ \ \ =pi \ (128-128/7 - 0)
\ \ \ =(768pi)/7 , as before