# 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

Volume

#### Explanation:

If you imagine an almost infinitesimally thin vertical line of thickness

#delta A ~~"width" xx "height" = ydeltax = f(x)deltax#

If we then rotated this infinitesimally thin vertical line about

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

Similarly if we rotate about

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

So for this problem we have:

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=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

# 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

# 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 (

# 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