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.
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(-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.