How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y=5x^2#, y=5x revolved about the x-axis?

In order to use shells, we need to take our representative slice parallel to the axis of rotation. So we will take a horizontal slice at a value of #y# and of thickness #dy#

Below is a partial picture. The curves intersect at #(0,0)# and #1,5)#.

The horizontal slice is shown in black and the radius of revolution is the dotted black line. Its length is #y#.

The volume of a representative shell is #2pirh*"thickness"#

In this problem, #"thickness" = dy# and #r = y#

#h# is the greater #x# value - the lesser #x# value.
The greater #x# value is the one on the right.

We need to rewrite the equation #y = 5x^2# so that we have #x# as a function of #y#. (#y# is the variable we'll be using for the integration.)

#x = sqrt(y/5)# (We ignore the negative square root because #x# it positive in this set-up.)

The lesser #x# is on the left where #x = y/5#

So the reprentative shell has volume

#2piy(sqrt(y/5)-y/5) dy#

#y# varies from #0# to #5#, so the solid has volume

#V = int_0^5 pi y(sqrt(y/5)-y/5) dy#

Details omitted

# = (5pi)/3#