Question

What is the surface area of the solid created by revolving `f(x) =2x+5 , x in [1,2]` around the x axis?

Answer 1

We need to calculate the area of te curved surface and the two end circles.

Explanation:

The end circles have radii `f(1)=7` and `f(2)=9` respectively and so have areas `49pi` and `81pi`.

The are of the curved surface may be calculated by considering an infinitesimal ring of radius `f(x)` between the values `x` and `x+dx`, which will have thickness `dl=sqrt(df^2(x)+dx^2)` and therefore `Area=2pif(x)dl`. The total curved area will then be `\int_1^2 2pif(x)dl`
Now, `dl=sqrt(df^2(x)+dx^2)=dxsqrt(((df(x))/dx)^2+1)=dxsqrt(2^2+1)`
`=dxsqrt5`.
So we have to evaluate the integral
`\int_1^2 dx 2pi f(x) sqrt5=2sqrt5pi [2x^2/2+5x]_1^2=2sqrt5pi [14-6]`
`=16sqrt5pi`.

The total area is then `=(16sqrt5+49+81)pi=(16sqrt5+130)pi`.