# What is the volume of the solid produced by revolving f(x)=x^2-x+1, x in [1,3] around x=1?

Mar 30, 2017

$\frac{32 \pi}{3}$

#### Explanation:

The given function is a parabola opening upwards, as shown in the figure. The portion to revolved lies between x=1 and 3 and the axis of revolution is x=1

Now consider an elementary strip of thickness $\Delta y$ and length (x-1) at a distance y from the x-axis. On revolving it around x=1, the volume of the solid so generated would be $\pi {\left(x - 1\right)}^{2} \Delta y$. Pl see attached figure.

The volume of the whole solid so generated would be

${\int}_{y = 1}^{y = 7} \pi {\left(x - 1\right)}^{2} \mathrm{dy}$

From the given eq y= ${x}^{2} - x + 1$ it would be dy= (2x-1)dx. Using this substitution, the volume integral would become

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

=$\pi {\int}_{1}^{3} \left(2 {x}^{3} - 5 {x}^{2} + 4 x - 1\right) \mathrm{dx}$

=$\pi {\left[2 {x}^{4} / 4 - 5 {x}^{3} / 3 + 4 {x}^{2} / 2 - x\right]}_{1}^{3}$

=$\pi \left[\frac{81}{2} - \frac{135}{3} + 18 - 3 - \left(\frac{1}{2} - \frac{5}{3} + 2 - 1\right)\right]$

=$\pi \left[\frac{81}{2} - \frac{135}{3} + 15 - \frac{3}{2} + \frac{5}{3}\right]$

=$\pi \left[39 + 15 - \frac{130}{3}\right]$ =$\frac{32 \pi}{3}$

Apr 1, 2017

The region is not fully specified. I'll answer both possible questions.

#### Explanation:

$f \left(x\right) = {x}^{2} - x + 1$ for $x \in \left[1 , 3\right]$ around the line $x = 1$.

Upper Region
If the region is above the parabola and below $y = 7$, then we have:
The region bounded by the functions is shown below in red.
If we take a representative slice parallel to the axis of revolution we have a slice at $x$ and thickness of $\mathrm{dx}$.

When we revolve the slice, we get a cylindrical shell with volume

$2 \pi r h \cdot \text{thickness}$

In this case $r = x - 1$
(Because we're revolving around $x = 1$ the radius is the distance between the $x$ value of the slice and $1$.)

and $h = 7 - \left({x}^{2} - x + 1\right)$
(The height of the slice is the upper $y$ value minus the lower.)

The thickness is $\mathrm{dx}$

The volume of the representative shell is: $2 \pi \left(x - 1\right) \left(7 - \left({x}^{2} - x + 1\right)\right) \mathrm{dx}$

We will be integrating with respect to $x$, so we observe that $x$ varies from $1$ to $3$.

The volume of the solid is
$V = {\int}_{1}^{3} 2 \pi \left(x - 1\right) \left(7 - \left({x}^{2} - x + 1\right)\right) \mathrm{dx}$

$= 2 \pi {\int}_{1}^{3} \left(x - 1\right) \left(7 - \left({x}^{2} - x + 1\right)\right) \mathrm{dx}$

$= 2 \pi \left(\frac{16}{3}\right) = \frac{32 \pi}{3}$

LowerRegion
If the region is belowthe parabola and above the $x$-axis, then we have:
The region bounded by the functions is shown below in red.
If we take a representative slice parallel to the axis of revolution we have a slice at $x$ and thickness of $\mathrm{dx}$.

When we revolve the slice, we get a cylindrical shell with volume

$2 \pi r h \cdot \text{thickness}$

In this case $r = x - 1$
(Because we're revolving around $x = 1$ the radius is the distance between the $x$ value of the slice and $1$.)

and $h = \left({x}^{2} - x + 1\right)$
(The height of the slice is the upper $y$ value minus the lower. value, which is $0$)

The thickness is $\mathrm{dx}$

The volume of the representative shell is: $2 \pi \left(x - 1\right) \left({x}^{2} - x + 1\right) \mathrm{dx}$

We will be integrating with respect to $x$, so we observe that $x$ varies from $1$ to $3$.

The volume of the solid is
$V = {\int}_{1}^{3} 2 \pi \left(x - 1\right) \left({x}^{2} - x + 1\right) \mathrm{dx}$

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

$= 2 \pi \left(\frac{26}{3}\right) = \frac{52 \pi}{3}$