# Find the volume of the solid generated by revolving about the line x=-1, the region bounded by the curves y=-x^2+4x-3 and y=0.

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Let,

EQ.1: `y=-x^2+4x-3` and EQ.2: `y =0`

Solve the intersection points of EQ.1 and EQ.2. Note that we already have a value of y which is zero. So, substitute y=0 to EQ.1.

`0 = -x^2+4x-3`

`0 = (-x+1)(x-3)`

Set both factors to zero and solve for x.

`-x+1 = 0 ` and `x-3 = 0`

` x = 1 ` `x = 3`

Hence, EQ.1 and EQ.2 intersect at (1,0) and (3,0).

So our, graph is:

If we revolve the bounded region of the two equations about x=-1, to determine the volume generated, we may use the cylindrical shell method.

`V = int_a^b A(x)dx= 2piint_a^brhdx`

where A(x) - is the area of the cylinder

r - is the radius or distance between the axis of revolution and the cylindrical shell

h - is the height of the cylindrical shell

a & b - are the x-coordinates of the intersection of two curves

Base on the graph, we have

`r =x - (-1) = x+1`

`h = y_(EQ.1) - y_(EQ.2) = -x^2+4x-3-0 = -x^2+4x-3`

Then, substitute this to the formula of volume.

`V=2pi int_1^3(x+1)(-x^2+4x-3)dx=2piint_1^3(-x^3+3x^2+x-3)dx`

`V = 2pi(-x^4/4+x^3+x^2/2-3x)|_1^3`

`V= 2pi[(-3^4/4 + 3^3 + 3^2/2-3(3)) - (-1^4/4+1^3+1^2/2-3(1))]`

`V = 2pi[(-81/4 + 27 + 9/2 - 9) - (-1/4+1+1/2-3)]`

`V = 2pi (9/4- (-7/4)) = 2pi(4) = 8pi`

**Answer: Volume is `8pi` cubic units.**