# 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 $\mathrm{dx}$, and height is the area given ($3 x - {x}^{2}$)... then what is the radius? I don't know what that would be (is it $x$?) What I have, plugged into equation: $V = 2 \setminus \pi \setminus {\int}_{0}^{3} \left[r \cdot \left(3 - {x}^{2}\right) \mathrm{dx}\right]$

Jun 6, 2018

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 $\left(x + 1\right)$ is the radius.

$V = 2 \pi {\int}_{0}^{3} \left(x + 1\right) \left(3 x - {x}^{2}\right) \cdot \mathrm{dx}$

$V = 2 \pi {\int}_{0}^{3} \left(3 {x}^{2} - {x}^{3} + 3 x - {x}^{2}\right) \cdot \mathrm{dx} = 2 \pi {\int}_{0}^{3} \left(2 {x}^{2} - {x}^{3} + 3 x\right) \cdot \mathrm{dx}$

$2 \pi \cdot {\left[\frac{2}{3} \cdot {x}^{3} - \frac{1}{4} \cdot {x}^{4} + \frac{3}{2} \cdot {x}^{2}\right]}_{0}^{3} = \frac{45 \pi}{2}$

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