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

##### 2 Answers

#### 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

The volume of the whole solid so generated would be

From the given eq y=

=

=

=

=

=

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

#### Explanation:

**Upper Region**

If the region is above the parabola and below

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

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

In this case

(Because we're revolving around

and

(The height of the slice is the upper

The thickness is

The volume of the representative shell is:

We will be integrating with respect to

The volume of the solid is

# = 2pi int_1^3 (x-1)(7-(x^2-x+1))dx#

# = 2pi(16/3) = (32pi)/3#

**LowerRegion**

If the region is belowthe parabola and above the

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

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

In this case

(Because we're revolving around

and

(The height of the slice is the upper

The thickness is

The volume of the representative shell is:

We will be integrating with respect to

The volume of the solid is

# = 2pi int_1^3 (x-1)(x^2-x+1)dx#

# = 2pi(26/3) = (52pi)/3#