Question

Find the volume using cylindrical shells? (Enclosed by x-axis and parabola `y=3x-x^2`, revolved about `x=-1`)

So far, I have the bounds at 0 and 3.
...I think I've forgotten how to apply shells, exactly.

Assuming thickness is `dx`, and height is the area given (`3x-x^2`)... then what is the radius? I don't know what that would be (is it `x`?)

What I have, plugged into equation:
`V=2\pi\int_0^3[r*(3-x^2)dx]`

Answer 1

`color(blue)[V=2piint_0^3(x+1)(3x-x^2)*dx=(45pi)/2`

Explanation:

In cylindrical shell method the slice should be parallel to the axis of revolution.

In your question the area bounded revolving around `x=-1` so our integral will given by :

`color(red)[V=2piint_0^3(x+1)(3x-x^2)*dx`

where `(x+1)` is the radius.

`V=2piint_0^3(x+1)(3x-x^2)*dx`

`V=2piint_0^3(3x^2-x^3+3x-x^2)*dx=2piint_0^3(2x^2-x^3+3x)*dx`

`2pi*[2/3*x^3-1/4*x^4+3/2*x^2]_0^3=(45pi)/2`

show below the region (shaded) revolving around `x=-1`: