Question

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?

Answer 1

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)`