How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=6x^2#, #y=6sqrtx# rotated about the y-axis?

1 Answer
Truong-Son N.
Jul 5, 2015

The normal revolution method calls for:

#V = pir^2h = sum_(a=a_0)^b pi(r(x))^2Deltax = piint_a^b (r(x))^2dx#

where one stacks circle analogs of varying radius #r(x)# across an orthogonal interval #Deltax#.

In contrast, the shell method calls for a volume formula as such:

#V = int_a^b 2pixr(x)dx#

where the way #r(x)# varies determines the height and shape of the revolved solid. Here, #Deltax# is instead the thickness of the shell and #x# is its radius. This is sometimes easier, especially if you are rotating about the y-axis instead of the x-axis.

Let's see how this looks.

#y = 6x^2#:
graph{6x^2 [-2, 2, -1, 1]}

#y = 6sqrtx#:
graph{6sqrtx [-2, 2, -2, 6]}

If you layer these graphs on top of each other, you can see that they intersect to form a "stretched lemon" of sorts. Let's find where they intersect to determine our #Deltax#.

Besides #x = 0#:

#6x^2 = 6sqrtx#
#x^2 = sqrtx#
#x^4 = x#
#x^3 = 1#
#x = 1#

Thus, the practical interval is #[0, 1]#, as you can see here.

So, taking the area between the two curves as the difference between the top #(6sqrtx)# and bottom #(6x^2)# curves and using it as #r(x)#:

#int_0^1 2pixr(x)dx#

#= 2piint_0^1 x[6sqrtx - 6x^2]dx#

#= 12piint_0^1 x[sqrtx - x^2]dx#

#= 12piint_0^1 x^(3/2) - x^3dx#

#= 12pi [2/5x^(5/2) - x^4/4]|_(0)^(1)#

#= 12pi [(2/5(1)^(5/2) - (1)^4/4) - (2/5(0)^(5/2) - (0)^4/4)]#

#= 12pi (2/5 - 1/4)#

#= 12pi (8/20 - 5/20)#

#= 12pi (3/20)#

#= 36/20 pi#

#= color(blue)(9/5 pi ~~ 5.6549 "u"^3)#

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