Question

What is the surface area of the solid created by revolving `f(x) = x^2-3x+2 , x in [3,4]` around the x axis?

Answer 1

`SA = 103 \ unit^2` (3sf)

Explanation:

The surface area of a solid created by revolving `y=f(x)` about `Ox` is given by:

`SA = 2pi \ int_(x=alpha)^(x=beta) \ f(x) \ sqrt(1+(f'(x))^2) \ dx`

So we have;

`\ \ \ \ \ \ f(x) = x^2 - 3x + 2`
`:. f'(x) = 2x-3`

And so the Surface Area is;

`SA = 2pi \ int_(3)^(4) \ (x^2-3x+2) \ sqrt(1+(2x-3)^2) \ dx`
`\ \ \ \ \ = 2pi \ int_(3)^(4) \ (x^2-3x+2) \ sqrt(1+(4x^2-12x+9)) \ dx`
`\ \ \ \ \ = 2pi \ int_(3)^(4) \ (x^2-3x+2) \ sqrt( 4x^2-12x+10 ) \ dx`

Although the integral can be established it is quite complex,so I will just quote the result

`SA = pi/32 (5sinh^-1(3)-45sqrt(10) -5sinh^-1(5) +235sqrt(26))`
`\ \ \ \ \ = 103.426839 ...`#